Explanation:
Okay, so I need to solve this equation: arccos x + arccos 2x + arccos 3x = π. And then find the cubic polynomial that x satisfies, with integer coefficients, and then find the smallest possible value of |a| + |b| + |c| + |d|. Alright, let me start by recalling some trigonometric identities and properties of inverse cosine functions.

First, I know that the range of arccos is [0, π], so each term arccos x, arccos 2x, arccos 3x must be in that interval. Therefore, the arguments of arccos must be in [-1, 1]. So, for arccos x to be defined, x must be in [-1, 1]. Similarly, 2x must be in [-1, 1], so x ∈ [-0.5, 0.5]. Similarly, 3x must be in [-1, 1], so x ∈ [-1/3, 1/3]. Therefore, combining all these, x must be in [-1/3, 1/3]. So x is between -1/3 and 1/3. But since arccos is decreasing, and the sum of three arccos terms equals π, which is a positive number. Let me think about the possible values of x. If x is negative, then arccos x would be in (π/2, π), arccos 2x would be in (π/2, π) if 2x is negative, same with arccos 3x. But adding three angles each greater than π/2 would give more than 3π/2, which is more than π. So maybe x is positive? Let me check.

If x is positive, then arccos x, arccos 2x, arccos 3x are all in [0, π/2], since their arguments are positive and less than or equal to 1. Wait, but 3x is up to 1 when x is 1/3. So arccos(3x) would be arccos(1) = 0 when x = 1/3. So if x is positive, then arccos x is between 0 and π/2, arccos 2x is between 0 and π/2, arccos 3x is between 0 and π/2. So adding three angles between 0 and π/2. The maximum sum would be 3*(π/2) = 3π/2, but the equation says the sum is π. So maybe x is positive? Let me check if x can be negative. Suppose x is negative. Then arccos x is in (π/2, π), arccos 2x is in (π/2, π) if 2x is between -1 and 0, same with arccos 3x. So adding three angles each between π/2 and π would give a sum between 3π/2 and 3π. But the equation says the sum is π, which is less than 3π/2. Therefore, x cannot be negative. Therefore, x must be positive. Therefore, x ∈ (0, 1/3].

So x is between 0 and 1/3. So that's helpful. So we can focus on x in (0, 1/3].

Now, the equation is arccos x + arccos 2x + arccos 3x = π. Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. So maybe I can use some trigonometric identities here. Let me think. If I take cos of both sides? Let me try that.

But cos(α + β + γ) = cos π = -1. But expanding cos(α + β + γ) might be complicated. Alternatively, maybe I can use the identity for cos(α + β) = cos π - γ. Wait, since α + β + γ = π, then α + β = π - γ. Therefore, cos(α + β) = cos(π - γ) = -cos γ. So cos(α + β) = -cos γ. Let me compute cos(α + β) using the formula:

cos(α + β) = cos α cos β - sin α sin β. And we know that cos(α + β) = -cos γ. Therefore:

cos α cos β - sin α sin β = -cos γ.

But cos α = x, cos β = 2x, cos γ = 3x. Therefore:

x * 2x - sin α sin β = -3x.

So 2x² - sin α sin β = -3x.

Therefore, sin α sin β = 2x² + 3x.

Hmm. Now, I need to find expressions for sin α and sin β. Since α = arccos x, then sin α = sqrt(1 - x²). Similarly, sin β = sqrt(1 - (2x)^2) = sqrt(1 - 4x²). Therefore:

sqrt(1 - x²) * sqrt(1 - 4x²) = 2x² + 3x.

So that's an equation in x. Let me square both sides to eliminate the square roots. But before that, let me note that squaring can introduce extraneous solutions, so I need to check later.

So squaring both sides:

(1 - x²)(1 - 4x²) = (2x² + 3x)^2.

Let me compute both sides.

Left side: (1 - x²)(1 - 4x²) = 1*(1 - 4x²) - x²*(1 - 4x²) = 1 - 4x² - x² + 4x^4 = 1 - 5x² + 4x^4.

Right side: (2x² + 3x)^2 = (2x²)^2 + 2*(2x²)*(3x) + (3x)^2 = 4x^4 + 12x^3 + 9x².

Therefore, equation becomes:

1 - 5x² + 4x^4 = 4x^4 + 12x^3 + 9x².

Subtracting left side from both sides:

0 = 4x^4 + 12x^3 + 9x² - (1 - 5x² + 4x^4) = 4x^4 + 12x^3 + 9x² -1 + 5x² -4x^4.

Simplify:

4x^4 -4x^4 cancels. Then 12x^3 + 9x² +5x² -1 = 12x^3 +14x² -1.

Therefore, 12x^3 +14x² -1 = 0.

Wait, so the equation reduces to 12x³ +14x² -1 = 0? Let me check my steps again to make sure.

Original equation after squaring:

Left side: 1 -5x² +4x^4.

Right side: 4x^4 +12x^3 +9x².

Subtract left side from right side:

(4x^4 +12x^3 +9x²) - (1 -5x² +4x^4) = 4x^4 -4x^4 +12x^3 +9x² +5x² -1 = 12x^3 +14x² -1. Yes, that's correct. Therefore, 12x³ +14x² -1 = 0. So that's a cubic equation. Wait, but the problem says that x satisfies a cubic polynomial of the form ax³ +bx² +cx +d = 0. So here, we have 12x³ +14x² +0x -1 = 0. Therefore, a=12, b=14, c=0, d=-1. Then |a| + |b| + |c| + |d| = 12 +14 +0 +1 = 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." So maybe there is a cubic polynomial with smaller coefficients? Or perhaps I made a mistake in the derivation?

Wait, but let me check if this equation is correct. Let me verify with an example. Suppose x is a solution. Let me check if squaring introduced extraneous solutions. Let me suppose that x satisfies 12x³ +14x² -1 = 0. Let me try to find approximate value of x. Let me try x=0. Let me plug into 12x³ +14x² -1: -1. At x=0.1: 12*(0.001) +14*(0.01) -1 = 0.012 + 0.14 -1 = -0.848. At x=0.2: 12*(0.008) +14*(0.04) -1 = 0.096 + 0.56 -1 = -0.344. At x=0.25: 12*(0.015625) +14*(0.0625) -1 = 0.1875 + 0.875 -1 = 1.0625 -1 = 0.0625. So between 0.2 and 0.25, the function crosses zero. Let me check x=0.23: 12*(0.23)^3 +14*(0.23)^2 -1. Let me compute 0.23^2 = 0.0529, 0.23^3 ≈ 0.012167. Then 12*0.012167 ≈ 0.146, 14*0.0529 ≈ 0.7406. Total ≈ 0.146 + 0.7406 -1 ≈ 0.8866 -1 ≈ -0.1134. At x=0.24: 0.24^2=0.0576, 0.24^3=0.013824. 12*0.013824≈0.1659, 14*0.0576≈0.8064. Total ≈ 0.1659 + 0.8064 -1 ≈ 0.9723 -1 ≈ -0.0277. At x=0.245: 0.245^2 ≈ 0.0600, 0.245^3 ≈ 0.0148. 12*0.0148 ≈ 0.1776, 14*0.06 ≈ 0.84. Total ≈ 0.1776 + 0.84 -1 ≈ 1.0176 -1 ≈ 0.0176. So between 0.24 and 0.245, the root is. So approximate value around 0.242. Let me check if this x satisfies the original equation arccos x + arccos 2x + arccos 3x = π. Let me compute each term. Let me take x ≈ 0.242.

Compute arccos(0.242): Let me recall that arccos(0.242) ≈ 1.33 radians (since cos(π/2) = 0, cos(π/3)=0.5, so 0.242 is between π/3 and π/2. Let me compute cos(1.33): cos(1.33) ≈ cos(76 degrees) ≈ 0.2419. So arccos(0.242) ≈ 1.33 radians.

arccos(2x) = arccos(0.484). Let me compute that. cos(π/3) = 0.5, so 0.484 is slightly less than 0.5, so arccos(0.484) ≈ π/3 + a little. Let me compute cos(1.047) = 0.5. Let me compute cos(1.05 radians): cos(1.05) ≈ 0.525. Wait, that's higher. Wait, maybe I need to think differently. Let me use calculator-like approach. Let me note that arccos(0.484) is approximately... Let me use the approximation. Let me take θ = arccos(0.484). Then cos θ = 0.484. Let me use Taylor series or something? Alternatively, since I know that cos(π/3) = 0.5, so θ is slightly larger than π/3. Let me compute θ ≈ π/3 + Δ. Then cos(π/3 + Δ) ≈ cos π/3 cos Δ - sin π/3 sin Δ ≈ 0.5*(1 - Δ²/2) - (√3/2)*(Δ). Setting this equal to 0.484:

0.5*(1 - Δ²/2) - (√3/2)*Δ ≈ 0.484.

So 0.5 - 0.25Δ² - (√3/2)Δ ≈ 0.484.

Then 0.5 - 0.484 ≈ 0.25Δ² + (√3/2)Δ.

So 0.016 ≈ 0.25Δ² + 0.866Δ.

Multiply both sides by 4: 0.064 ≈ Δ² + 3.464Δ.

So Δ² + 3.464Δ - 0.064 ≈ 0. Solving quadratic equation:

Δ = [-3.464 ± sqrt(3.464² + 4*0.064)] / 2.

Compute discriminant: 12 + 0.256 ≈ 12.256. sqrt(12.256) ≈ 3.5. Therefore, Δ ≈ [-3.464 + 3.5]/2 ≈ 0.036/2 ≈ 0.018. So θ ≈ π/3 + 0.018 ≈ 1.047 + 0.018 ≈ 1.065 radians. Therefore, arccos(0.484) ≈ 1.065 radians.

Similarly, arccos(3x) = arccos(0.726). Let me compute that. cos(π/4) ≈ 0.707, so 0.726 is slightly larger. So arccos(0.726) is slightly less than π/4. Let me compute θ = arccos(0.726). Let me use similar approach. Let me take θ = π/4 - Δ. Then cos(π/4 - Δ) = cos π/4 cos Δ + sin π/4 sin Δ ≈ (√2/2)(1 - Δ²/2) + (√2/2)(Δ). Setting equal to 0.726:

(√2/2)(1 - Δ²/2 + Δ) ≈ 0.726.

Compute √2/2 ≈ 0.7071. So:

0.7071*(1 + Δ - Δ²/2) ≈ 0.726.

Divide both sides by 0.7071:

1 + Δ - Δ²/2 ≈ 0.726 / 0.7071 ≈ 1.026.

Therefore:

Δ - Δ²/2 ≈ 1.026 - 1 = 0.026.

Assuming Δ is small, Δ² is negligible, so Δ ≈ 0.026. Therefore, θ ≈ π/4 - 0.026 ≈ 0.785 - 0.026 ≈ 0.759 radians.

Therefore, sum of angles: arccos x + arccos 2x + arccos 3x ≈ 1.33 + 1.065 + 0.759 ≈ 3.154 radians. π is approximately 3.1416. So the sum is approximately 3.154, which is slightly larger than π. So there's a discrepancy here. Which suggests that maybe my approximate value of x is not accurate enough, or perhaps there is an error in my derivation?

Wait, but I squared the equation, which can introduce extraneous solutions. So maybe the solution to 12x³ +14x² -1 = 0 is not actually a solution to the original equation? Let me check with more accurate calculation.

Alternatively, maybe I made a mistake in the derivation. Let me check again.

Original equation: arccos x + arccos 2x + arccos 3x = π.

Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Then α + β = π - γ. Taking cosine of both sides: cos(α + β) = cos(π - γ) = -cos γ. Then expanding cos(α + β) = cos α cos β - sin α sin β = -cos γ. Then substituting cos α = x, cos β = 2x, cos γ = 3x. Therefore:

x * 2x - sin α sin β = -3x => 2x² - sin α sin β = -3x => sin α sin β = 2x² + 3x. Then sin α = sqrt(1 - x²), sin β = sqrt(1 - 4x²). Therefore:

sqrt(1 - x²) * sqrt(1 - 4x²) = 2x² + 3x. Then squaring both sides:

(1 - x²)(1 - 4x²) = (2x² + 3x)^2. Which gives 1 -5x² +4x^4 = 4x^4 +12x^3 +9x². Then simplifying gives 0 = 12x^3 +14x² -1. So that seems correct. Therefore, the equation is correct. However, when I checked with approximate x ≈ 0.242, the sum of angles was slightly larger than π. Maybe my approximation was not precise enough? Let me try to compute more accurately.

Alternatively, maybe I need to check if the equation 12x³ +14x² -1 = 0 has a root in (0, 1/3). Let me check x=1/3 ≈ 0.333. Let me compute 12*(1/3)^3 +14*(1/3)^2 -1 = 12*(1/27) +14*(1/9) -1 = (12/27) + (14/9) -1 = (4/9) + (14/9) -1 = 18/9 -1 = 2 -1 = 1. So at x=1/3, the value is 1. At x=0.24, as before, the value was approximately -0.0277. So there is a root between 0.24 and 0.333. Wait, but earlier when I tried x≈0.242, the sum of angles was approx 3.154, which is larger than π≈3.1416. Maybe the actual root is slightly less than 0.242? Let me try x=0.24. Let me compute the left side of the original equation.

Let me take x=0.24. Then:

arccos(0.24) ≈ 1.33 radians (as before). arccos(0.48) ≈ let me check with calculator. Let me recall that cos(1.047) = 0.5, so 0.48 is less than 0.5, so arccos(0.48) is slightly more than π/3 ≈ 1.047. Let me use calculator-like computation. Let me use the inverse cosine approximation. Let me use the formula for arccos(y). Alternatively, use a calculator. Wait, since I don't have a calculator here, maybe I can use more accurate estimation.

Alternatively, maybe I can use the equation 12x³ +14x² -1 = 0. Let me try to find a better approximation. Let me take x=0.24. Let me compute 12*(0.24)^3 +14*(0.24)^2 -1. 0.24^2 = 0.0576, 0.24^3 = 0.013824. Then 12*0.013824 = 0.165888, 14*0.0576 = 0.8064. Total: 0.165888 + 0.8064 -1 = 0.972288 -1 = -0.027712. So f(0.24) ≈ -0.0277. At x=0.245, f(x)=12*(0.245)^3 +14*(0.245)^2 -1. Let me compute 0.245^2 = 0.060025, 0.245^3 = 0.245*0.060025 ≈ 0.014706. Then 12*0.014706 ≈ 0.17647, 14*0.060025 ≈ 0.84035. Total ≈ 0.17647 + 0.84035 -1 ≈ 1.0168 -1 = 0.0168. So f(0.245) ≈ 0.0168. Therefore, the root is between 0.24 and 0.245. Let me use linear approximation. Let me denote x1=0.24, f(x1)= -0.0277; x2=0.245, f(x2)=0.0168. The difference in x is 0.005, difference in f is 0.0168 - (-0.0277)=0.0445. To find Δx such that f(x1 + Δx)=0. So Δx ≈ (0 - f(x1)) * (x2 - x1)/ (f(x2)-f(x1)) ) = (0.0277)*(0.005)/0.0445 ≈ 0.000138 / 0.0445 ≈ 0.0031. Therefore, approximate root at x ≈ 0.24 + 0.0031 ≈ 0.2431. Let me check x=0.2431.

Compute f(0.2431):

First compute x=0.2431.

x² ≈ (0.2431)^2 ≈ 0.0591.

x³ ≈ 0.2431 * 0.0591 ≈ 0.01436.

Then 12x³ ≈ 12*0.01436 ≈ 0.1723.

14x² ≈ 14*0.0591 ≈ 0.8274.

Total f(x) ≈ 0.1723 + 0.8274 -1 ≈ 0.9997 -1 ≈ -0.0003. Close to zero. So x≈0.2431 is a root. Let me take x≈0.243. Let me compute the sum arccos x + arccos 2x + arccos 3x with x≈0.243.

Compute arccos(0.243). Let me use calculator-like estimation. Let me recall that cos(π/2) = 0, cos(π/3)=0.5. Let me use a better approach. Let me use the value of x≈0.243. Let me use the inverse cosine approximation. Alternatively, since I know that cos(1.33) ≈ 0.2419 as before. So arccos(0.243) is slightly less than 1.33 radians. Let me compute cos(1.33) ≈ 0.2419. Let me compute cos(1.32). Let me use the Taylor series around π/2? Maybe not. Alternatively, use linear approximation. Let me take θ = 1.33 - Δ, such that cos(θ) = 0.243. Let me take θ = 1.33 - Δ. Then cos(θ) ≈ cos(1.33) + sin(1.33)*Δ. Wait, derivative of cos(θ) is -sin(θ). So cos(θ) ≈ cos(1.33) - sin(1.33)*Δ. Let me compute sin(1.33). Since 1.33 radians is approx 76 degrees. sin(1.33) ≈ sin(π/2 - 0.23) ≈ cos(0.23) ≈ 0.974. Wait, more accurately, sin(1.33) ≈ sin(π/2 - 0.23) = cos(0.23) ≈ 0.974. Therefore, cos(θ) ≈ 0.2419 - 0.974*Δ = 0.243. Therefore, 0.2419 - 0.974Δ = 0.243 => -0.974Δ = 0.0011 => Δ ≈ -0.0011 / 0.974 ≈ -0.0011. Therefore, θ ≈ 1.33 - (-0.0011) = 1.33 + 0.0011 ≈ 1.3311 radians. Therefore, arccos(0.243) ≈ 1.3311 radians.

Similarly, arccos(2x) = arccos(0.486). Let me compute that. Let me take θ = arccos(0.486). Let me use similar approach. Let me note that cos(π/3) = 0.5, so θ is slightly larger than π/3. Let me take θ = π/3 + Δ. Then cos(θ) ≈ cos(π/3) - sin(π/3)*Δ ≈ 0.5 - (√3/2)*Δ. Set equal to 0.486:

0.5 - (√3/2)*Δ ≈ 0.486 => (√3/2)*Δ ≈ 0.014 => Δ ≈ 0.014 * 2 / √3 ≈ 0.028 / 1.732 ≈ 0.016. Therefore, θ ≈ π/3 + 0.016 ≈ 1.047 + 0.016 ≈ 1.063 radians.

Similarly, arccos(3x) = arccos(0.729). Let me compute that. Let me note that cos(π/4) ≈ 0.707, so θ is slightly less than π/4. Let me take θ = π/4 - Δ. Then cos(θ) ≈ cos(π/4) + sin(π/4)*Δ ≈ √2/2 + (√2/2)*Δ. Set equal to 0.729:

√2/2 + (√2/2)*Δ ≈ 0.729. Since √2/2 ≈ 0.7071, then:

0.7071 + 0.7071*Δ ≈ 0.729 => 0.7071*Δ ≈ 0.729 - 0.7071 ≈ 0.0219 => Δ ≈ 0.0219 / 0.7071 ≈ 0.031. Therefore, θ ≈ π/4 - 0.031 ≈ 0.785 - 0.031 ≈ 0.754 radians.

Therefore, sum of angles ≈ 1.3311 + 1.063 + 0.754 ≈ 1.3311 + 1.063 = 2.3941 + 0.754 ≈ 3.1481 radians. Which is approximately π + 0.0065 radians. So very close to π. Considering the approximated nature of my calculations, this seems acceptable. Therefore, the equation 12x³ +14x² -1 = 0 is correct, and the root is approximately 0.243, which satisfies the original equation approximately. Therefore, the cubic equation is correct.

But the problem says "x satisfies a cubic polynomial of the form ax³ +bx² +cx +d = 0, where a, b, c, d are integers, and a ≠ 0. Find the smallest possible value of |a| + |b| + |c| + |d|."

So the equation we found is 12x³ +14x² +0x -1 = 0. Therefore, coefficients are a=12, b=14, c=0, d=-1. Then |a| + |b| + |c| + |d| = 12 +14 +0 +1 = 27. However, the problem says "smallest possible value". Maybe there is another cubic polynomial with smaller coefficients? For example, maybe the equation can be simplified or there is a common factor?

But 12x³ +14x² -1. Let me check if this polynomial can be factored. Let me try rational roots. The possible rational roots are ±1, ±1/2, ±1/3, ±1/4, ±1/6, ±1/12. Let me test them.

For x=1: 12 +14 -1 = 25 ≠0.

x=-1: -12 +14 -1=1≠0.

x=1/2: 12*(1/8) +14*(1/4) -1 = 1.5 + 3.5 -1 = 4 ≠0.

x=-1/2: 12*(-1/2)^3 +14*(-1/2)^2 -1 = 12*(-1/8) +14*(1/4) -1 = -1.5 + 3.5 -1 = 1 ≠0.

x=1/3: 12*(1/27) +14*(1/9) -1 = 4/9 +14/9 -1 = 18/9 -1 = 2 -1 =1≠0.

x=-1/3: 12*(-1/3)^3 +14*(-1/3)^2 -1 = 12*(-1/27) +14*(1/9) -1 = -4/9 +14/9 -1 = 10/9 -1 = 1/9 ≠0.

x=1/4: 12*(1/64) +14*(1/16) -1 = 12/64 +14/16 -1 = 3/16 +7/8 -1 = 3/16 +14/16 -16/16 = 1/16 ≠0.

x=-1/4: 12*(-1/4)^3 +14*(-1/4)^2 -1 = 12*(-1/64) +14*(1/16) -1 = -12/64 +14/16 -1 = -3/16 +7/8 -1 = (-3/16 +14/16) -1 = 11/16 -1 = -5/16 ≠0.

x=1/6: 12*(1/216) +14*(1/36) -1 = 12/216 +14/36 -1 = 1/18 +7/18 -1 = 8/18 -1 = 4/9 -1 = -5/9 ≠0.

x=-1/6: 12*(-1/6)^3 +14*(-1/6)^2 -1 = 12*(-1/216) +14*(1/36) -1 = -1/18 +7/18 -1 = 6/18 -1 = 1/3 -1 = -2/3 ≠0.

x=1/12: 12*(1/12)^3 +14*(1/12)^2 -1 = 12*(1/1728) +14*(1/144) -1 = 1/144 + 14/144 -1 = 15/144 -1 = 5/48 -1 = -43/48 ≠0.

So no rational roots. Therefore, the polynomial 12x³ +14x² -1 is irreducible over rationals? Therefore, it cannot be factored into lower-degree polynomials with integer coefficients. Therefore, this is the minimal polynomial? Wait, but maybe there is another equation? Wait, but we derived this equation through valid steps, so unless there was an error in derivation, this should be the correct equation. However, the problem says "the cubic polynomial of the form...", so maybe there is another cubic polynomial with smaller coefficients? For example, if we multiply by something or something else?

Wait, but the coefficients are 12, 14, 0, -1. The sum |a| + |b| + |c| + |d| is 12 +14 +0 +1 = 27. But maybe there is another way to derive the equation with smaller coefficients? Let me think.

Alternatively, maybe I made a mistake in the earlier steps. Let me check again.

Original equation: arccos x + arccos 2x + arccos 3x = π.

Let me try another approach. Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Let me consider taking two arccos terms and combining them. For example, take α + β = π - γ. Then, as before, cos(α + β) = -cos γ. Which led us to the equation. Alternatively, maybe take another combination. For example, take α + γ = π - β. Then cos(α + γ) = -cos β. Let me try that.

So cos(α + γ) = cos α cos γ - sin α sin γ = -cos β. Which is x*(3x) - sqrt(1 -x²)*sqrt(1 -9x²) = -2x. Therefore:

3x² - sqrt(1 -x²)sqrt(1 -9x²) = -2x.

Rearranged:

sqrt(1 -x²)sqrt(1 -9x²) = 3x² + 2x.

Then square both sides:

(1 -x²)(1 -9x²) = (3x² + 2x)^2.

Compute left side: 1 -9x² -x² +9x^4 = 1 -10x² +9x^4.

Right side: 9x^4 + 12x^3 +4x².

Therefore, equation:

1 -10x² +9x^4 = 9x^4 +12x^3 +4x².

Subtract left side:

0 = 12x^3 +14x² -1. Same equation as before! So same result. Therefore, regardless of which combination I take, I get the same equation. Therefore, the equation is correct. Therefore, the cubic is indeed 12x³ +14x² -1 = 0. Therefore, the coefficients are 12, 14, 0, -1. Therefore, the sum is 27. However, the problem says "smallest possible value of |a| + |b| + |c| + |d|". Maybe there is another cubic polynomial with smaller coefficients? For example, if we can factor out something? But since the polynomial is irreducible over integers, as we saw, there are no rational roots, so it cannot be factored into lower-degree polynomials with integer coefficients. Therefore, this is the minimal polynomial. Therefore, the answer is 27? But maybe I need to check if there is another approach that leads to a different cubic equation?

Alternatively, maybe I need to consider another identity. Let me think. Suppose I take all three arccos terms and use some other identity. For example, take the sum α + β + γ = π. Let me consider taking cos(α + β + γ) = cos π = -1. But expanding cos(α + β + γ) is complicated. Let me try:

cos(α + β + γ) = cos α cos β cos γ - cos α sin β sin γ - sin α cos β sin γ - sin α sin β cos γ. Wait, that seems complicated. Alternatively, use the identity:

cos(α + β + γ) = cos α cos β cos γ - cos α sin β sin γ - sin α cos β sin γ - sin α sin β cos γ. But since α + β + γ = π, then cos(α + β + γ) = -1. Therefore:

-1 = cos α cos β cos γ - cos α sin β sin γ - sin α cos β sin γ - sin α sin β cos γ.

But this seems too complicated. Let me substitute the known values. cos α = x, cos β = 2x, cos γ = 3x. Therefore:

-1 = x*(2x)*(3x) - x*sin β sin γ - sin α * 2x * sin γ - sin α sin β * 3x.

Simplify:

-1 = 6x³ - x sin β sin γ - 2x sin α sin γ - 3x sin α sin β.

But from earlier, we had sin α sin β = 2x² + 3x. Also, we can find expressions for sin α sin γ and sin β sin γ?

Alternatively, maybe this approach is more complex. Let me see. Let me note that we already have an equation from before, which is 12x³ +14x² -1 = 0. Therefore, unless there is a miscalculation, this is the correct equation. Therefore, the answer is 27. However, the problem says "smallest possible value", so maybe there is a different cubic equation? For example, if we consider that maybe the original equation has multiple solutions, and the minimal polynomial is different? But no, the equation we derived is the minimal polynomial for the root x. Since the polynomial is irreducible, it's the minimal polynomial. Therefore, the answer is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, if we multiply by something? But the coefficients are already integers, and the question says "a cubic polynomial of the form...", so it's possible that there is another cubic equation with smaller coefficients. Wait, but how?

Alternatively, maybe I made a mistake in squaring steps. Let me check again. When I squared both sides of sqrt(1 -x²)sqrt(1 -4x²) = 2x² + 3x, I obtained (1 -x²)(1 -4x²) = (2x² + 3x)^2. But maybe there is a mistake here? Let me check:

Left side: sqrt(A) * sqrt(B) = sqrt(AB). Squaring gives AB. Right side: (2x² + 3x)^2. So that is correct. Therefore, the equation is correct. Therefore, the cubic equation is correct. Therefore, the answer is 27. However, I recall that sometimes when you square equations, you can get extraneous solutions, but in this case, we have only one real root in the interval (0, 1/3), so the equation is correct. Therefore, the answer is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, if we can write the equation as 12x³ +14x² -1 = 0, but maybe there is a multiple of this equation with smaller coefficients? For example, if we divide by something? But since coefficients are integers, and the polynomial is irreducible, we can't divide by anything. Therefore, 12x³ +14x² -1 is the minimal polynomial. Therefore, the answer is 27. However, I need to check if the problem says "smallest possible value of |a| + |b| + |c| + |d|", so maybe there is another cubic equation with different coefficients? For example, if we consider that maybe there is a different approach leading to another cubic equation?

Alternatively, maybe I can use another identity. Let me think. Suppose I take two arccos terms and combine them. For example, take arccos x + arccos 2x = π - arccos 3x. Then take cosine of both sides:

cos(arccos x + arccos 2x) = cos(π - arccos 3x) = -cos(arccos 3x) = -3x.

So cos(arccos x + arccos 2x) = -3x. Let me compute the left side using the formula:

cos(A + B) = cos A cos B - sin A sin B. Which is the same as before. Which gives x*2x - sqrt(1 -x²)sqrt(1 -4x²) = -3x. Which leads to the same equation as before. Therefore, same result. Therefore, no difference.

Alternatively, maybe take another combination. For example, take arccos 2x + arccos 3x = π - arccos x. Then take cosine of both sides:

cos(arccos 2x + arccos 3x) = cos(π - arccos x) = -cos(arccos x) = -x.

Compute left side:

cos(arccos 2x + arccos 3x) = cos(arccos 2x)cos(arccos 3x) - sin(arccos 2x)sin(arccos 3x) = 2x * 3x - sqrt(1 -4x²)sqrt(1 -9x²) = 6x² - sqrt(1 -4x²)sqrt(1 -9x²). Therefore:

6x² - sqrt(1 -4x²)sqrt(1 -9x²) = -x.

Rearranged:

sqrt(1 -4x²)sqrt(1 -9x²) = 6x² + x.

Square both sides:

(1 -4x²)(1 -9x²) = (6x² + x)^2.

Left side: 1 -9x² -4x² +36x^4 = 1 -13x² +36x^4.

Right side: 36x^4 +12x^3 +x².

Equation:

1 -13x² +36x^4 = 36x^4 +12x^3 +x².

Subtract left side:

0 = 12x^3 +14x² -1. Same equation again! So same result. Therefore, regardless of the approach, we get the same cubic equation. Therefore, the answer is 27. Therefore, the minimal value is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, if we can write the equation as something else. Let me think. Suppose we have 12x³ +14x² -1 = 0. Let me check if this can be written as a multiple of another cubic with smaller coefficients. For example, if we divide by something? But since coefficients are integers, and the polynomial is irreducible, we can't. Therefore, 12x³ +14x² -1 is the minimal polynomial. Therefore, the answer is 27. However, I need to check if the problem allows for different forms. For example, maybe the equation can be written with different signs? For example, if we multiply by -1: -12x³ -14x² +1 = 0. Then |a| + |b| + |c| + |d| is still 12 +14 +0 +1 = 27. Same value. Therefore, no improvement.

Alternatively, maybe there is a different cubic equation? For example, if I consider that maybe there was a miscalculation in the squaring step. Let me check again:

Original equation after squaring: (1 -x²)(1 -4x²) = (2x² + 3x)^2.

Left side: 1 -5x² +4x^4.

Right side: 4x^4 +12x^3 +9x².

Subtracting left side from right side: 0 = 12x^3 +14x² -1. Correct. Therefore, equation is correct. Therefore, the answer is 27. Therefore, I think the answer is 27. However, I recall that sometimes problems like this might have a different cubic equation if you consider more steps. Let me think again.

Alternatively, maybe I need to consider the original equation and use more identities. For example, take the three arccos terms and use some trigonometric identities. Let me think. Suppose I let α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Let me consider taking tan of both sides? Not sure. Alternatively, use sine of both sides? Let me try.

Take sin(α + β + γ) = sin π = 0. But expanding sin(α + β + γ) is complicated. Let me try:

sin(α + β + γ) = sin α cos β cos γ + cos α sin β cos γ + cos α cos β sin γ - sin α sin β sin γ = 0.

Substituting the known values:

sin α = sqrt(1 -x²), sin β = sqrt(1 -4x²), sin γ = sqrt(1 -9x²).

cos α = x, cos β = 2x, cos γ = 3x.

Therefore:

sqrt(1 -x²)*2x*3x + x*sqrt(1 -4x²)*3x + x*2x*sqrt(1 -9x²) - sqrt(1 -x²)*sqrt(1 -4x²)*sqrt(1 -9x²) = 0.

Simplify:

6x² sqrt(1 -x²) + 3x² sqrt(1 -4x²) + 2x² sqrt(1 -9x²) - sqrt(1 -x²)sqrt(1 -4x²)sqrt(1 -9x²) = 0.

This seems even more complicated. I don't think this approach is helpful. Therefore, I think the previous method is the correct one, leading to the cubic equation 12x³ +14x² -1 = 0. Therefore, the answer is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, if I consider that maybe there was a miscalculation in the squaring step. Let me check again:

Original equation after squaring:

(1 -x²)(1 -4x²) = (2x² + 3x)^2.

Left side: 1 -4x² -x² +4x^4 = 1 -5x² +4x^4.

Right side: 4x^4 + 12x^3 +9x².

Subtracting left side from right side: 0 = 12x^3 +14x² -1. Correct. Therefore, equation is correct. Therefore, the answer is 27. Therefore, I think the answer is 27. However, I need to check if the problem says "smallest possible value of |a| + |b| + |c| + |d|", and maybe there is another cubic equation with smaller coefficients. For example, if we consider that maybe the equation can be written as something else. For example, if we multiply by something? But since the polynomial is irreducible, I don't think so. Therefore, I think the answer is 27. However, I recall that sometimes problems like this have a different answer. Let me check with another approach.

Alternatively, suppose I let y = x. Then the equation is arccos y + arccos 2y + arccos 3y = π. Let me consider taking the first two arccos terms and combining them. Let me denote θ = arccos y + arccos 2y. Then θ = π - arccos 3y. Taking cosine of both sides:

cos θ = cos(π - arccos 3y) = -cos(arccos 3y) = -3y.

But θ = arccos y + arccos 2y. So cos θ = cos(arccos y + arccos 2y) = cos(arccos y)cos(arccos 2y) - sin(arccos y)sin(arccos 2y) = y * 2y - sqrt(1 - y²) * sqrt(1 - 4y²) = 2y² - sqrt(1 - y²)sqrt(1 - 4y²). Therefore:

2y² - sqrt(1 - y²)sqrt(1 - 4y²) = -3y.

Rearranged:

sqrt(1 - y²)sqrt(1 - 4y²) = 2y² + 3y.

Which is the same equation as before. Squaring gives the same cubic equation. Therefore, same result. Therefore, no difference. Therefore, I think the answer is indeed 27. Therefore, I think the answer is \boxed{27}.

**Final Answer**
\boxed{27}

To solve the equation:

$$
\arccos x + \arccos 2x + \arccos 3x = \pi,
$$

we begin by analyzing the domain of the inverse cosine function. Since the range of $\arccos$ is $[0, \pi]$, and the arguments of the inverse cosine functions must lie in $[-1, 1]$, we deduce that:

- $x \in [-1, 1]$
- $2x \in [-1, 1] \Rightarrow x \in [-0.5, 0.5]$
- $3x \in [-1, 1] \Rightarrow x \in [-1/3, 1/3]$

Thus, the domain of $x$ is $[-1/3, 1/3]$. However, since the sum of three arccosines equals $\pi$, and the arccos function is decreasing, we can infer that $x$ must be **positive**. If $x$ were negative, the sum of the arccosines would exceed $\pi$, which contradicts the given equation. Therefore, we restrict $x \in (0, 1/3]$.

---

### Step 1: Use Trigonometric Identities

Let:
- $\alpha = \arccos x$
- $\beta = \arccos 2x$
- $\gamma = \arccos 3x$

Then the given equation becomes:
$$
\alpha + \beta + \gamma = \pi
$$

From this, we can write:
$$
\alpha + \beta = \pi - \gamma
$$

Taking the cosine of both sides:
$$
\cos(\alpha + \beta) = \cos(\pi - \gamma) = -\cos \gamma
$$

Using the identity for $\cos(\alpha + \beta)$:
$$
\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
$$

Substituting known values:
- $\cos \alpha = x$
- $\cos \beta = 2x$
- $\cos \gamma = 3x$

We get:
$$
x \cdot 2x - \sin \alpha \sin \beta = -3x
\Rightarrow 2x^2 - \sin \alpha \sin \beta = -3x
\Rightarrow \sin \alpha \sin \beta = 2x^2 + 3x
$$

Now, using the identity $\sin^2 \theta = 1 - \cos^2 \theta$, we find:
- $\sin \alpha = \sqrt{1 - x^2}$
- $\sin \beta = \sqrt{1 - 4x^2}$

So:
$$
\sqrt{1 - x^2} \cdot \sqrt{1 - 4x^2} = 2x^2 + 3x
$$

---

### Step 2: Square Both Sides

Squaring both sides:
$$
(1 - x^2)(1 - 4x^2) = (2x^2 + 3x)^2
$$

Compute both sides:

- Left: $(1 - x^2)(1 - 4x^2) = 1 - 5x^2 + 4x^4$
- Right: $(2x^2 + 3x)^2 = 4x^4 + 12x^3 + 9x^2$

Set them equal:
$$
1 - 5x^2 + 4x^4 = 4x^4 + 12x^3 + 9x^2
$$

Subtracting the left side from the right:
$$
0 = 12x^3 + 14x^2 - 1
$$

So the cubic equation is:
$$
12x^3 + 14x^2 - 1 = 0
$$

---

### Step 3: Analyze the Polynomial

This is a cubic polynomial with integer coefficients:
$$
12x^3 + 14x^2 + 0x - 1 = 0
$$

We are asked to find the **smallest possible value of** $|a| + |b| + |c| + |d|$ where the polynomial is of the form $ax^3 + bx^2 + cx + d = 0$.

Here:
- $a = 12$
- $b = 14$
- $c = 0$
- $d = -1$

So:
$$
|a| + |b| + |c| + |d| = 12 + 14 + 0 + 1 = 27
$$

We also check whether this polynomial can be simplified or factored further. Since it has no rational roots (as verified by the Rational Root Theorem), and the coefficients are integers, this is the **minimal polynomial** with integer coefficients that the root $x$ satisfies.

---

### Final Answer

$$
\boxed{27}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: arccos x + arccos 2x + arccos 3x = π. And then find the cubic polynomial that x satisfies, with integer coefficients, and then find the smallest possible value of |a| + |b| + |c| + |d|. Alright, let me start by recalling some trigonometric identities and properties of inverse cosine functions.

First, I know that the range of arccos is [0, π], so each term arccos x, arccos 2x, arccos 3x must be in that interval. Therefore, the arguments of arccos must be in [-1, 1]. So, for arccos x to be defined, x must be in [-1, 1]. Similarly, 2x must be in [-1, 1], so x ∈ [-0.5, 0.5]. Similarly, 3x must be in [-1, 1], so x ∈ [-1/3, 1/3]. Therefore, combining all these, x must be in [-1/3, 1/3]. So x is between -1/3 and 1/3. But since arccos is decreasing, and the sum of three arccos terms equals π, which is a positive number. Let me think about the possible values of x. If x is negative, then arccos x would be in (π/2, π), arccos 2x would be in (π/2, π) if 2x is negative, same with arccos 3x. But adding three angles each greater than π/2 would give more than 3π/2, which is more than π. So maybe x is positive? Let me check.

If x is positive, then arccos x, arccos 2x, arccos 3x are all in [0, π/2], since their arguments are positive and less than or equal to 1. Wait, but 3x is up to 1 when x is 1/3. So arccos(3x) would be arccos(1) = 0 when x = 1/3. So if x is positive, then arccos x is between 0 and π/2, arccos 2x is between 0 and π/2, arccos 3x is between 0 and π/2. So adding three angles between 0 and π/2. The maximum sum would be 3*(π/2) = 3π/2, but the equation says the sum is π. So maybe x is positive? Let me check if x can be negative. Suppose x is negative. Then arccos x is in (π/2, π), arccos 2x is in (π/2, π) if 2x is between -1 and 0, same with arccos 3x. So adding three angles each between π/2 and π would give a sum between 3π/2 and 3π. But the equation says the sum is π, which is less than 3π/2. Therefore, x cannot be negative. Therefore, x must be positive. Therefore, x ∈ (0, 1/3].

So x is between 0 and 1/3. So that's helpful. So we can focus on x in (0, 1/3].

Now, the equation is arccos x + arccos 2x + arccos 3x = π. Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. So maybe I can use some trigonometric identities here. Let me think. If I take cos of both sides? Let me try that.

But cos(α + β + γ) = cos π = -1. But expanding cos(α + β + γ) might be complicated. Alternatively, maybe I can use the identity for cos(α + β) = cos π - γ. Wait, since α + β + γ = π, then α + β = π - γ. Therefore, cos(α + β) = cos(π - γ) = -cos γ. So cos(α + β) = -cos γ. Let me compute cos(α + β) using the formula:

cos(α + β) = cos α cos β - sin α sin β. And we know that cos(α + β) = -cos γ. Therefore:

cos α cos β - sin α sin β = -cos γ.

But cos α = x, cos β = 2x, cos γ = 3x. Therefore:

x * 2x - sin α sin β = -3x.

So 2x² - sin α sin β = -3x.

Therefore, sin α sin β = 2x² + 3x.

Hmm. Now, I need to find expressions for sin α and sin β. Since α = arccos x, then sin α = sqrt(1 - x²). Similarly, sin β = sqrt(1 - (2x)^2) = sqrt(1 - 4x²). Therefore:

sqrt(1 - x²) * sqrt(1 - 4x²) = 2x² + 3x.

So that's an equation in x. Let me square both sides to eliminate the square roots. But before that, let me note that squaring can introduce extraneous solutions, so I need to check later.

So squaring both sides:

(1 - x²)(1 - 4x²) = (2x² + 3x)^2.

Let me compute both sides.

Left side: (1 - x²)(1 - 4x²) = 1*(1 - 4x²) - x²*(1 - 4x²) = 1 - 4x² - x² + 4x^4 = 1 - 5x² + 4x^4.

Right side: (2x² + 3x)^2 = (2x²)^2 + 2*(2x²)*(3x) + (3x)^2 = 4x^4 + 12x^3 + 9x².

Therefore, equation becomes:

1 - 5x² + 4x^4 = 4x^4 + 12x^3 + 9x².

Subtracting left side from both sides:

0 = 4x^4 + 12x^3 + 9x² - (1 - 5x² + 4x^4) = 4x^4 + 12x^3 + 9x² -1 + 5x² -4x^4.

Simplify:

4x^4 -4x^4 cancels. Then 12x^3 + 9x² +5x² -1 = 12x^3 +14x² -1.

Therefore, 12x^3 +14x² -1 = 0.

Wait, so the equation reduces to 12x³ +14x² -1 = 0? Let me check my steps again to make sure.

Original equation after squaring:

Left side: 1 -5x² +4x^4.

Right side: 4x^4 +12x^3 +9x².

Subtract left side from right side:

(4x^4 +12x^3 +9x²) - (1 -5x² +4x^4) = 4x^4 -4x^4 +12x^3 +9x² +5x² -1 = 12x^3 +14x² -1. Yes, that's correct. Therefore, 12x³ +14x² -1 = 0. So that's a cubic equation. Wait, but the problem says that x satisfies a cubic polynomial of the form ax³ +bx² +cx +d = 0. So here, we have 12x³ +14x² +0x -1 = 0. Therefore, a=12, b=14, c=0, d=-1. Then |a| + |b| + |c| + |d| = 12 +14 +0 +1 = 27. But the problem says "Find the smallest possible value of |a| + |b| + |c| + |d|." So maybe there is a cubic polynomial with smaller coefficients? Or perhaps I made a mistake in the derivation?

Wait, but let me check if this equation is correct. Let me verify with an example. Suppose x is a solution. Let me check if squaring introduced extraneous solutions. Let me suppose that x satisfies 12x³ +14x² -1 = 0. Let me try to find approximate value of x. Let me try x=0. Let me plug into 12x³ +14x² -1: -1. At x=0.1: 12*(0.001) +14*(0.01) -1 = 0.012 + 0.14 -1 = -0.848. At x=0.2: 12*(0.008) +14*(0.04) -1 = 0.096 + 0.56 -1 = -0.344. At x=0.25: 12*(0.015625) +14*(0.0625) -1 = 0.1875 + 0.875 -1 = 1.0625 -1 = 0.0625. So between 0.2 and 0.25, the function crosses zero. Let me check x=0.23: 12*(0.23)^3 +14*(0.23)^2 -1. Let me compute 0.23^2 = 0.0529, 0.23^3 ≈ 0.012167. Then 12*0.012167 ≈ 0.146, 14*0.0529 ≈ 0.7406. Total ≈ 0.146 + 0.7406 -1 ≈ 0.8866 -1 ≈ -0.1134. At x=0.24: 0.24^2=0.0576, 0.24^3=0.013824. 12*0.013824≈0.1659, 14*0.0576≈0.8064. Total ≈ 0.1659 + 0.8064 -1 ≈ 0.9723 -1 ≈ -0.0277. At x=0.245: 0.245^2 ≈ 0.0600, 0.245^3 ≈ 0.0148. 12*0.0148 ≈ 0.1776, 14*0.06 ≈ 0.84. Total ≈ 0.1776 + 0.84 -1 ≈ 1.0176 -1 ≈ 0.0176. So between 0.24 and 0.245, the root is. So approximate value around 0.242. Let me check if this x satisfies the original equation arccos x + arccos 2x + arccos 3x = π. Let me compute each term. Let me take x ≈ 0.242.

Compute arccos(0.242): Let me recall that arccos(0.242) ≈ 1.33 radians (since cos(π/2) = 0, cos(π/3)=0.5, so 0.242 is between π/3 and π/2. Let me compute cos(1.33): cos(1.33) ≈ cos(76 degrees) ≈ 0.2419. So arccos(0.242) ≈ 1.33 radians.

arccos(2x) = arccos(0.484). Let me compute that. cos(π/3) = 0.5, so 0.484 is slightly less than 0.5, so arccos(0.484) ≈ π/3 + a little. Let me compute cos(1.047) = 0.5. Let me compute cos(1.05 radians): cos(1.05) ≈ 0.525. Wait, that's higher. Wait, maybe I need to think differently. Let me use calculator-like approach. Let me note that arccos(0.484) is approximately... Let me use the approximation. Let me take θ = arccos(0.484). Then cos θ = 0.484. Let me use Taylor series or something? Alternatively, since I know that cos(π/3) = 0.5, so θ is slightly larger than π/3. Let me compute θ ≈ π/3 + Δ. Then cos(π/3 + Δ) ≈ cos π/3 cos Δ - sin π/3 sin Δ ≈ 0.5*(1 - Δ²/2) - (√3/2)*(Δ). Setting this equal to 0.484:

0.5*(1 - Δ²/2) - (√3/2)*Δ ≈ 0.484.

So 0.5 - 0.25Δ² - (√3/2)Δ ≈ 0.484.

Then 0.5 - 0.484 ≈ 0.25Δ² + (√3/2)Δ.

So 0.016 ≈ 0.25Δ² + 0.866Δ.

Multiply both sides by 4: 0.064 ≈ Δ² + 3.464Δ.

So Δ² + 3.464Δ - 0.064 ≈ 0. Solving quadratic equation:

Δ = [-3.464 ± sqrt(3.464² + 4*0.064)] / 2.

Compute discriminant: 12 + 0.256 ≈ 12.256. sqrt(12.256) ≈ 3.5. Therefore, Δ ≈ [-3.464 + 3.5]/2 ≈ 0.036/2 ≈ 0.018. So θ ≈ π/3 + 0.018 ≈ 1.047 + 0.018 ≈ 1.065 radians. Therefore, arccos(0.484) ≈ 1.065 radians.

Similarly, arccos(3x) = arccos(0.726). Let me compute that. cos(π/4) ≈ 0.707, so 0.726 is slightly larger. So arccos(0.726) is slightly less than π/4. Let me compute θ = arccos(0.726). Let me use similar approach. Let me take θ = π/4 - Δ. Then cos(π/4 - Δ) = cos π/4 cos Δ + sin π/4 sin Δ ≈ (√2/2)(1 - Δ²/2) + (√2/2)(Δ). Setting equal to 0.726:

(√2/2)(1 - Δ²/2 + Δ) ≈ 0.726.

Compute √2/2 ≈ 0.7071. So:

0.7071*(1 + Δ - Δ²/2) ≈ 0.726.

Divide both sides by 0.7071:

1 + Δ - Δ²/2 ≈ 0.726 / 0.7071 ≈ 1.026.

Therefore:

Δ - Δ²/2 ≈ 1.026 - 1 = 0.026.

Assuming Δ is small, Δ² is negligible, so Δ ≈ 0.026. Therefore, θ ≈ π/4 - 0.026 ≈ 0.785 - 0.026 ≈ 0.759 radians.

Therefore, sum of angles: arccos x + arccos 2x + arccos 3x ≈ 1.33 + 1.065 + 0.759 ≈ 3.154 radians. π is approximately 3.1416. So the sum is approximately 3.154, which is slightly larger than π. So there's a discrepancy here. Which suggests that maybe my approximate value of x is not accurate enough, or perhaps there is an error in my derivation?

Wait, but I squared the equation, which can introduce extraneous solutions. So maybe the solution to 12x³ +14x² -1 = 0 is not actually a solution to the original equation? Let me check with more accurate calculation.

Alternatively, maybe I made a mistake in the derivation. Let me check again.

Original equation: arccos x + arccos 2x + arccos 3x = π.

Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Then α + β = π - γ. Taking cosine of both sides: cos(α + β) = cos(π - γ) = -cos γ. Then expanding cos(α + β) = cos α cos β - sin α sin β = -cos γ. Then substituting cos α = x, cos β = 2x, cos γ = 3x. Therefore:

x * 2x - sin α sin β = -3x => 2x² - sin α sin β = -3x => sin α sin β = 2x² + 3x. Then sin α = sqrt(1 - x²), sin β = sqrt(1 - 4x²). Therefore:

sqrt(1 - x²) * sqrt(1 - 4x²) = 2x² + 3x. Then squaring both sides:

(1 - x²)(1 - 4x²) = (2x² + 3x)^2. Which gives 1 -5x² +4x^4 = 4x^4 +12x^3 +9x². Then simplifying gives 0 = 12x^3 +14x² -1. So that seems correct. Therefore, the equation is correct. However, when I checked with approximate x ≈ 0.242, the sum of angles was slightly larger than π. Maybe my approximation was not precise enough? Let me try to compute more accurately.

Alternatively, maybe I need to check if the equation 12x³ +14x² -1 = 0 has a root in (0, 1/3). Let me check x=1/3 ≈ 0.333. Let me compute 12*(1/3)^3 +14*(1/3)^2 -1 = 12*(1/27) +14*(1/9) -1 = (12/27) + (14/9) -1 = (4/9) + (14/9) -1 = 18/9 -1 = 2 -1 = 1. So at x=1/3, the value is 1. At x=0.24, as before, the value was approximately -0.0277. So there is a root between 0.24 and 0.333. Wait, but earlier when I tried x≈0.242, the sum of angles was approx 3.154, which is larger than π≈3.1416. Maybe the actual root is slightly less than 0.242? Let me try x=0.24. Let me compute the left side of the original equation.

Let me take x=0.24. Then:

arccos(0.24) ≈ 1.33 radians (as before). arccos(0.48) ≈ let me check with calculator. Let me recall that cos(1.047) = 0.5, so 0.48 is less than 0.5, so arccos(0.48) is slightly more than π/3 ≈ 1.047. Let me use calculator-like computation. Let me use the inverse cosine approximation. Let me use the formula for arccos(y). Alternatively, use a calculator. Wait, since I don't have a calculator here, maybe I can use more accurate estimation.

Alternatively, maybe I can use the equation 12x³ +14x² -1 = 0. Let me try to find a better approximation. Let me take x=0.24. Let me compute 12*(0.24)^3 +14*(0.24)^2 -1. 0.24^2 = 0.0576, 0.24^3 = 0.013824. Then 12*0.013824 = 0.165888, 14*0.0576 = 0.8064. Total: 0.165888 + 0.8064 -1 = 0.972288 -1 = -0.027712. So f(0.24) ≈ -0.0277. At x=0.245, f(x)=12*(0.245)^3 +14*(0.245)^2 -1. Let me compute 0.245^2 = 0.060025, 0.245^3 = 0.245*0.060025 ≈ 0.014706. Then 12*0.014706 ≈ 0.17647, 14*0.060025 ≈ 0.84035. Total ≈ 0.17647 + 0.84035 -1 ≈ 1.0168 -1 = 0.0168. So f(0.245) ≈ 0.0168. Therefore, the root is between 0.24 and 0.245. Let me use linear approximation. Let me denote x1=0.24, f(x1)= -0.0277; x2=0.245, f(x2)=0.0168. The difference in x is 0.005, difference in f is 0.0168 - (-0.0277)=0.0445. To find Δx such that f(x1 + Δx)=0. So Δx ≈ (0 - f(x1)) * (x2 - x1)/ (f(x2)-f(x1)) ) = (0.0277)*(0.005)/0.0445 ≈ 0.000138 / 0.0445 ≈ 0.0031. Therefore, approximate root at x ≈ 0.24 + 0.0031 ≈ 0.2431. Let me check x=0.2431.

Compute f(0.2431):

First compute x=0.2431.

x² ≈ (0.2431)^2 ≈ 0.0591.

x³ ≈ 0.2431 * 0.0591 ≈ 0.01436.

Then 12x³ ≈ 12*0.01436 ≈ 0.1723.

14x² ≈ 14*0.0591 ≈ 0.8274.

Total f(x) ≈ 0.1723 + 0.8274 -1 ≈ 0.9997 -1 ≈ -0.0003. Close to zero. So x≈0.2431 is a root. Let me take x≈0.243. Let me compute the sum arccos x + arccos 2x + arccos 3x with x≈0.243.

Compute arccos(0.243). Let me use calculator-like estimation. Let me recall that cos(π/2) = 0, cos(π/3)=0.5. Let me use a better approach. Let me use the value of x≈0.243. Let me use the inverse cosine approximation. Alternatively, since I know that cos(1.33) ≈ 0.2419 as before. So arccos(0.243) is slightly less than 1.33 radians. Let me compute cos(1.33) ≈ 0.2419. Let me compute cos(1.32). Let me use the Taylor series around π/2? Maybe not. Alternatively, use linear approximation. Let me take θ = 1.33 - Δ, such that cos(θ) = 0.243. Let me take θ = 1.33 - Δ. Then cos(θ) ≈ cos(1.33) + sin(1.33)*Δ. Wait, derivative of cos(θ) is -sin(θ). So cos(θ) ≈ cos(1.33) - sin(1.33)*Δ. Let me compute sin(1.33). Since 1.33 radians is approx 76 degrees. sin(1.33) ≈ sin(π/2 - 0.23) ≈ cos(0.23) ≈ 0.974. Wait, more accurately, sin(1.33) ≈ sin(π/2 - 0.23) = cos(0.23) ≈ 0.974. Therefore, cos(θ) ≈ 0.2419 - 0.974*Δ = 0.243. Therefore, 0.2419 - 0.974Δ = 0.243 => -0.974Δ = 0.0011 => Δ ≈ -0.0011 / 0.974 ≈ -0.0011. Therefore, θ ≈ 1.33 - (-0.0011) = 1.33 + 0.0011 ≈ 1.3311 radians. Therefore, arccos(0.243) ≈ 1.3311 radians.

Similarly, arccos(2x) = arccos(0.486). Let me compute that. Let me take θ = arccos(0.486). Let me use similar approach. Let me note that cos(π/3) = 0.5, so θ is slightly larger than π/3. Let me take θ = π/3 + Δ. Then cos(θ) ≈ cos(π/3) - sin(π/3)*Δ ≈ 0.5 - (√3/2)*Δ. Set equal to 0.486:

0.5 - (√3/2)*Δ ≈ 0.486 => (√3/2)*Δ ≈ 0.014 => Δ ≈ 0.014 * 2 / √3 ≈ 0.028 / 1.732 ≈ 0.016. Therefore, θ ≈ π/3 + 0.016 ≈ 1.047 + 0.016 ≈ 1.063 radians.

Similarly, arccos(3x) = arccos(0.729). Let me compute that. Let me note that cos(π/4) ≈ 0.707, so θ is slightly less than π/4. Let me take θ = π/4 - Δ. Then cos(θ) ≈ cos(π/4) + sin(π/4)*Δ ≈ √2/2 + (√2/2)*Δ. Set equal to 0.729:

√2/2 + (√2/2)*Δ ≈ 0.729. Since √2/2 ≈ 0.7071, then:

0.7071 + 0.7071*Δ ≈ 0.729 => 0.7071*Δ ≈ 0.729 - 0.7071 ≈ 0.0219 => Δ ≈ 0.0219 / 0.7071 ≈ 0.031. Therefore, θ ≈ π/4 - 0.031 ≈ 0.785 - 0.031 ≈ 0.754 radians.

Therefore, sum of angles ≈ 1.3311 + 1.063 + 0.754 ≈ 1.3311 + 1.063 = 2.3941 + 0.754 ≈ 3.1481 radians. Which is approximately π + 0.0065 radians. So very close to π. Considering the approximated nature of my calculations, this seems acceptable. Therefore, the equation 12x³ +14x² -1 = 0 is correct, and the root is approximately 0.243, which satisfies the original equation approximately. Therefore, the cubic equation is correct.

But the problem says "x satisfies a cubic polynomial of the form ax³ +bx² +cx +d = 0, where a, b, c, d are integers, and a ≠ 0. Find the smallest possible value of |a| + |b| + |c| + |d|."

So the equation we found is 12x³ +14x² +0x -1 = 0. Therefore, coefficients are a=12, b=14, c=0, d=-1. Then |a| + |b| + |c| + |d| = 12 +14 +0 +1 = 27. However, the problem says "smallest possible value". Maybe there is another cubic polynomial with smaller coefficients? For example, maybe the equation can be simplified or there is a common factor?

But 12x³ +14x² -1. Let me check if this polynomial can be factored. Let me try rational roots. The possible rational roots are ±1, ±1/2, ±1/3, ±1/4, ±1/6, ±1/12. Let me test them.

For x=1: 12 +14 -1 = 25 ≠0.

x=-1: -12 +14 -1=1≠0.

x=1/2: 12*(1/8) +14*(1/4) -1 = 1.5 + 3.5 -1 = 4 ≠0.

x=-1/2: 12*(-1/2)^3 +14*(-1/2)^2 -1 = 12*(-1/8) +14*(1/4) -1 = -1.5 + 3.5 -1 = 1 ≠0.

x=1/3: 12*(1/27) +14*(1/9) -1 = 4/9 +14/9 -1 = 18/9 -1 = 2 -1 =1≠0.

x=-1/3: 12*(-1/3)^3 +14*(-1/3)^2 -1 = 12*(-1/27) +14*(1/9) -1 = -4/9 +14/9 -1 = 10/9 -1 = 1/9 ≠0.

x=1/4: 12*(1/64) +14*(1/16) -1 = 12/64 +14/16 -1 = 3/16 +7/8 -1 = 3/16 +14/16 -16/16 = 1/16 ≠0.

x=-1/4: 12*(-1/4)^3 +14*(-1/4)^2 -1 = 12*(-1/64) +14*(1/16) -1 = -12/64 +14/16 -1 = -3/16 +7/8 -1 = (-3/16 +14/16) -1 = 11/16 -1 = -5/16 ≠0.

x=1/6: 12*(1/216) +14*(1/36) -1 = 12/216 +14/36 -1 = 1/18 +7/18 -1 = 8/18 -1 = 4/9 -1 = -5/9 ≠0.

x=-1/6: 12*(-1/6)^3 +14*(-1/6)^2 -1 = 12*(-1/216) +14*(1/36) -1 = -1/18 +7/18 -1 = 6/18 -1 = 1/3 -1 = -2/3 ≠0.

x=1/12: 12*(1/12)^3 +14*(1/12)^2 -1 = 12*(1/1728) +14*(1/144) -1 = 1/144 + 14/144 -1 = 15/144 -1 = 5/48 -1 = -43/48 ≠0.

So no rational roots. Therefore, the polynomial 12x³ +14x² -1 is irreducible over rationals? Therefore, it cannot be factored into lower-degree polynomials with integer coefficients. Therefore, this is the minimal polynomial? Wait, but maybe there is another equation? Wait, but we derived this equation through valid steps, so unless there was an error in derivation, this should be the correct equation. However, the problem says "the cubic polynomial of the form...", so maybe there is another cubic polynomial with smaller coefficients? For example, if we multiply by something or something else?

Wait, but the coefficients are 12, 14, 0, -1. The sum |a| + |b| + |c| + |d| is 12 +14 +0 +1 = 27. But maybe there is another way to derive the equation with smaller coefficients? Let me think.

Alternatively, maybe I made a mistake in the earlier steps. Let me check again.

Original equation: arccos x + arccos 2x + arccos 3x = π.

Let me try another approach. Let me denote α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Let me consider taking two arccos terms and combining them. For example, take α + β = π - γ. Then, as before, cos(α + β) = -cos γ. Which led us to the equation. Alternatively, maybe take another combination. For example, take α + γ = π - β. Then cos(α + γ) = -cos β. Let me try that.

So cos(α + γ) = cos α cos γ - sin α sin γ = -cos β. Which is x*(3x) - sqrt(1 -x²)*sqrt(1 -9x²) = -2x. Therefore:

3x² - sqrt(1 -x²)sqrt(1 -9x²) = -2x.

Rearranged:

sqrt(1 -x²)sqrt(1 -9x²) = 3x² + 2x.

Then square both sides:

(1 -x²)(1 -9x²) = (3x² + 2x)^2.

Compute left side: 1 -9x² -x² +9x^4 = 1 -10x² +9x^4.

Right side: 9x^4 + 12x^3 +4x².

Therefore, equation:

1 -10x² +9x^4 = 9x^4 +12x^3 +4x².

Subtract left side:

0 = 12x^3 +14x² -1. Same equation as before! So same result. Therefore, regardless of which combination I take, I get the same equation. Therefore, the equation is correct. Therefore, the cubic is indeed 12x³ +14x² -1 = 0. Therefore, the coefficients are 12, 14, 0, -1. Therefore, the sum is 27. However, the problem says "smallest possible value of |a| + |b| + |c| + |d|". Maybe there is another cubic polynomial with smaller coefficients? For example, if we can factor out something? But since the polynomial is irreducible over integers, as we saw, there are no rational roots, so it cannot be factored into lower-degree polynomials with integer coefficients. Therefore, this is the minimal polynomial. Therefore, the answer is 27? But maybe I need to check if there is another approach that leads to a different cubic equation?

Alternatively, maybe I need to consider another identity. Let me think. Suppose I take all three arccos terms and use some other identity. For example, take the sum α + β + γ = π. Let me consider taking cos(α + β + γ) = cos π = -1. But expanding cos(α + β + γ) is complicated. Let me try:

cos(α + β + γ) = cos α cos β cos γ - cos α sin β sin γ - sin α cos β sin γ - sin α sin β cos γ. Wait, that seems complicated. Alternatively, use the identity:

cos(α + β + γ) = cos α cos β cos γ - cos α sin β sin γ - sin α cos β sin γ - sin α sin β cos γ. But since α + β + γ = π, then cos(α + β + γ) = -1. Therefore:

-1 = cos α cos β cos γ - cos α sin β sin γ - sin α cos β sin γ - sin α sin β cos γ.

But this seems too complicated. Let me substitute the known values. cos α = x, cos β = 2x, cos γ = 3x. Therefore:

-1 = x*(2x)*(3x) - x*sin β sin γ - sin α * 2x * sin γ - sin α sin β * 3x.

Simplify:

-1 = 6x³ - x sin β sin γ - 2x sin α sin γ - 3x sin α sin β.

But from earlier, we had sin α sin β = 2x² + 3x. Also, we can find expressions for sin α sin γ and sin β sin γ?

Alternatively, maybe this approach is more complex. Let me see. Let me note that we already have an equation from before, which is 12x³ +14x² -1 = 0. Therefore, unless there is a miscalculation, this is the correct equation. Therefore, the answer is 27. However, the problem says "smallest possible value", so maybe there is a different cubic equation? For example, if we consider that maybe the original equation has multiple solutions, and the minimal polynomial is different? But no, the equation we derived is the minimal polynomial for the root x. Since the polynomial is irreducible, it's the minimal polynomial. Therefore, the answer is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, if we multiply by something? But the coefficients are already integers, and the question says "a cubic polynomial of the form...", so it's possible that there is another cubic equation with smaller coefficients. Wait, but how?

Alternatively, maybe I made a mistake in squaring steps. Let me check again. When I squared both sides of sqrt(1 -x²)sqrt(1 -4x²) = 2x² + 3x, I obtained (1 -x²)(1 -4x²) = (2x² + 3x)^2. But maybe there is a mistake here? Let me check:

Left side: sqrt(A) * sqrt(B) = sqrt(AB). Squaring gives AB. Right side: (2x² + 3x)^2. So that is correct. Therefore, the equation is correct. Therefore, the cubic equation is correct. Therefore, the answer is 27. However, I recall that sometimes when you square equations, you can get extraneous solutions, but in this case, we have only one real root in the interval (0, 1/3), so the equation is correct. Therefore, the answer is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, if we can write the equation as 12x³ +14x² -1 = 0, but maybe there is a multiple of this equation with smaller coefficients? For example, if we divide by something? But since coefficients are integers, and the polynomial is irreducible, we can't divide by anything. Therefore, 12x³ +14x² -1 is the minimal polynomial. Therefore, the answer is 27. However, I need to check if the problem says "smallest possible value of |a| + |b| + |c| + |d|", so maybe there is another cubic equation with different coefficients? For example, if we consider that maybe there is a different approach leading to another cubic equation?

Alternatively, maybe I can use another identity. Let me think. Suppose I take two arccos terms and combine them. For example, take arccos x + arccos 2x = π - arccos 3x. Then take cosine of both sides:

cos(arccos x + arccos 2x) = cos(π - arccos 3x) = -cos(arccos 3x) = -3x.

So cos(arccos x + arccos 2x) = -3x. Let me compute the left side using the formula:

cos(A + B) = cos A cos B - sin A sin B. Which is the same as before. Which gives x*2x - sqrt(1 -x²)sqrt(1 -4x²) = -3x. Which leads to the same equation as before. Therefore, same result. Therefore, no difference.

Alternatively, maybe take another combination. For example, take arccos 2x + arccos 3x = π - arccos x. Then take cosine of both sides:

cos(arccos 2x + arccos 3x) = cos(π - arccos x) = -cos(arccos x) = -x.

Compute left side:

cos(arccos 2x + arccos 3x) = cos(arccos 2x)cos(arccos 3x) - sin(arccos 2x)sin(arccos 3x) = 2x * 3x - sqrt(1 -4x²)sqrt(1 -9x²) = 6x² - sqrt(1 -4x²)sqrt(1 -9x²). Therefore:

6x² - sqrt(1 -4x²)sqrt(1 -9x²) = -x.

Rearranged:

sqrt(1 -4x²)sqrt(1 -9x²) = 6x² + x.

Square both sides:

(1 -4x²)(1 -9x²) = (6x² + x)^2.

Left side: 1 -9x² -4x² +36x^4 = 1 -13x² +36x^4.

Right side: 36x^4 +12x^3 +x².

Equation:

1 -13x² +36x^4 = 36x^4 +12x^3 +x².

Subtract left side:

0 = 12x^3 +14x² -1. Same equation again! So same result. Therefore, regardless of the approach, we get the same cubic equation. Therefore, the answer is 27. Therefore, the minimal value is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, if we can write the equation as something else. Let me think. Suppose we have 12x³ +14x² -1 = 0. Let me check if this can be written as a multiple of another cubic with smaller coefficients. For example, if we divide by something? But since coefficients are integers, and the polynomial is irreducible, we can't. Therefore, 12x³ +14x² -1 is the minimal polynomial. Therefore, the answer is 27. However, I need to check if the problem allows for different forms. For example, maybe the equation can be written with different signs? For example, if we multiply by -1: -12x³ -14x² +1 = 0. Then |a| + |b| + |c| + |d| is still 12 +14 +0 +1 = 27. Same value. Therefore, no improvement.

Alternatively, maybe there is a different cubic equation? For example, if I consider that maybe there was a miscalculation in the squaring step. Let me check again:

Original equation after squaring: (1 -x²)(1 -4x²) = (2x² + 3x)^2.

Left side: 1 -5x² +4x^4.

Right side: 4x^4 +12x^3 +9x².

Subtracting left side from right side: 0 = 12x^3 +14x² -1. Correct. Therefore, equation is correct. Therefore, the answer is 27. Therefore, I think the answer is 27. However, I recall that sometimes problems like this might have a different cubic equation if you consider more steps. Let me think again.

Alternatively, maybe I need to consider the original equation and use more identities. For example, take the three arccos terms and use some trigonometric identities. Let me think. Suppose I let α = arccos x, β = arccos 2x, γ = arccos 3x. Then α + β + γ = π. Let me consider taking tan of both sides? Not sure. Alternatively, use sine of both sides? Let me try.

Take sin(α + β + γ) = sin π = 0. But expanding sin(α + β + γ) is complicated. Let me try:

sin(α + β + γ) = sin α cos β cos γ + cos α sin β cos γ + cos α cos β sin γ - sin α sin β sin γ = 0.

Substituting the known values:

sin α = sqrt(1 -x²), sin β = sqrt(1 -4x²), sin γ = sqrt(1 -9x²).

cos α = x, cos β = 2x, cos γ = 3x.

Therefore:

sqrt(1 -x²)*2x*3x + x*sqrt(1 -4x²)*3x + x*2x*sqrt(1 -9x²) - sqrt(1 -x²)*sqrt(1 -4x²)*sqrt(1 -9x²) = 0.

Simplify:

6x² sqrt(1 -x²) + 3x² sqrt(1 -4x²) + 2x² sqrt(1 -9x²) - sqrt(1 -x²)sqrt(1 -4x²)sqrt(1 -9x²) = 0.

This seems even more complicated. I don't think this approach is helpful. Therefore, I think the previous method is the correct one, leading to the cubic equation 12x³ +14x² -1 = 0. Therefore, the answer is 27. However, I need to check if there is a different cubic equation with smaller coefficients. For example, if I consider that maybe there was a miscalculation in the squaring step. Let me check again:

Original equation after squaring:

(1 -x²)(1 -4x²) = (2x² + 3x)^2.

Left side: 1 -4x² -x² +4x^4 = 1 -5x² +4x^4.

Right side: 4x^4 + 12x^3 +9x².

Subtracting left side from right side: 0 = 12x^3 +14x² -1. Correct. Therefore, equation is correct. Therefore, the answer is 27. Therefore, I think the answer is 27. However, I need to check if the problem says "smallest possible value of |a| + |b| + |c| + |d|", and maybe there is another cubic equation with smaller coefficients. For example, if we consider that maybe the equation can be written as something else. For example, if we multiply by something? But since the polynomial is irreducible, I don't think so. Therefore, I think the answer is 27. However, I recall that sometimes problems like this have a different answer. Let me check with another approach.

Alternatively, suppose I let y = x. Then the equation is arccos y + arccos 2y + arccos 3y = π. Let me consider taking the first two arccos terms and combining them. Let me denote θ = arccos y + arccos 2y. Then θ = π - arccos 3y. Taking cosine of both sides:

cos θ = cos(π - arccos 3y) = -cos(arccos 3y) = -3y.

But θ = arccos y + arccos 2y. So cos θ = cos(arccos y + arccos 2y) = cos(arccos y)cos(arccos 2y) - sin(arccos y)sin(arccos 2y) = y * 2y - sqrt(1 - y²) * sqrt(1 - 4y²) = 2y² - sqrt(1 - y²)sqrt(1 - 4y²). Therefore:

2y² - sqrt(1 - y²)sqrt(1 - 4y²) = -3y.

Rearranged:

sqrt(1 - y²)sqrt(1 - 4y²) = 2y² + 3y.

Which is the same equation as before. Squaring gives the same cubic equation. Therefore, same result. Therefore, no difference. Therefore, I think the answer is indeed 27. Therefore, I think the answer is \boxed{27}.

**Final Answer**
\boxed{27}

To solve the equation:

$$
\arccos x + \arccos 2x + \arccos 3x = \pi,
$$

we begin by analyzing the domain of the inverse cosine function. Since the range of $\arccos$ is $[0, \pi]$, and the arguments of the inverse cosine functions must lie in $[-1, 1]$, we deduce that:

- $x \in [-1, 1]$
- $2x \in [-1, 1] \Rightarrow x \in [-0.5, 0.5]$
- $3x \in [-1, 1] \Rightarrow x \in [-1/3, 1/3]$

Thus, the domain of $x$ is $[-1/3, 1/3]$. However, since the sum of three arccosines equals $\pi$, and the arccos function is decreasing, we can infer that $x$ must be **positive**. If $x$ were negative, the sum of the arccosines would exceed $\pi$, which contradicts the given equation. Therefore, we restrict $x \in (0, 1/3]$.

---

### Step 1: Use Trigonometric Identities

Let:
- $\alpha = \arccos x$
- $\beta = \arccos 2x$
- $\gamma = \arccos 3x$

Then the given equation becomes:
$$
\alpha + \beta + \gamma = \pi
$$

From this, we can write:
$$
\alpha + \beta = \pi - \gamma
$$

Taking the cosine of both sides:
$$
\cos(\alpha + \beta) = \cos(\pi - \gamma) = -\cos \gamma
$$

Using the identity for $\cos(\alpha + \beta)$:
$$
\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
$$

Substituting known values:
- $\cos \alpha = x$
- $\cos \beta = 2x$
- $\cos \gamma = 3x$

We get:
$$
x \cdot 2x - \sin \alpha \sin \beta = -3x
\Rightarrow 2x^2 - \sin \alpha \sin \beta = -3x
\Rightarrow \sin \alpha \sin \beta = 2x^2 + 3x
$$

Now, using the identity $\sin^2 \theta = 1 - \cos^2 \theta$, we find:
- $\sin \alpha = \sqrt{1 - x^2}$
- $\sin \beta = \sqrt{1 - 4x^2}$

So:
$$
\sqrt{1 - x^2} \cdot \sqrt{1 - 4x^2} = 2x^2 + 3x
$$

---

### Step 2: Square Both Sides

Squaring both sides:
$$
(1 - x^2)(1 - 4x^2) = (2x^2 + 3x)^2
$$

Compute both sides:

- Left: $(1 - x^2)(1 - 4x^2) = 1 - 5x^2 + 4x^4$
- Right: $(2x^2 + 3x)^2 = 4x^4 + 12x^3 + 9x^2$

Set them equal:
$$
1 - 5x^2 + 4x^4 = 4x^4 + 12x^3 + 9x^2
$$

Subtracting the left side from the right:
$$
0 = 12x^3 + 14x^2 - 1
$$

So the cubic equation is:
$$
12x^3 + 14x^2 - 1 = 0
$$

---

### Step 3: Analyze the Polynomial

This is a cubic polynomial with integer coefficients:
$$
12x^3 + 14x^2 + 0x - 1 = 0
$$

We are asked to find the **smallest possible value of** $|a| + |b| + |c| + |d|$ where the polynomial is of the form $ax^3 + bx^2 + cx + d = 0$.

Here:
- $a = 12$
- $b = 14$
- $c = 0$
- $d = -1$

So:
$$
|a| + |b| + |c| + |d| = 12 + 14 + 0 + 1 = 27
$$

We also check whether this polynomial can be simplified or factored further. Since it has no rational roots (as verified by the Rational Root Theorem), and the coefficients are integers, this is the **minimal polynomial** with integer coefficients that the root $x$ satisfies.

---

### Final Answer

$$
\boxed{27}
$$

Solution:
From the equation $\arccos x + \arccos 2x + \arccos 3x = \pi,$ $\arccos x + \arccos 2x = \pi - \arccos 3x,$ so
\[\cos (\arccos x + \arccos 2x) = \cos (\pi - \arccos 3x).\]From the angle addition formula, the left-hand side becomes
\begin{align*}
\cos (\arccos x + \arccos 2x) &= \cos (\arccos x) \cos (\arccos 2x) - \sin (\arccos x) \sin (\arccos 2x) \\
&= (x)(2x) - (\sqrt{1 - x^2})(\sqrt{1 - 4x^2}) \\
&= 2x^2 - \sqrt{(1 - x^2)(1 - 4x^2)}.
\end{align*}The right-hand side becomes
\[\cos (\pi - \arccos 3x) = -\cos (\arccos 3x) = -3x,\]so
\[2x^2 - \sqrt{(1 - x^2)(1 - 4x^2)} = -3x.\]Then $\sqrt{(1 - x^2)(1 - 4x^2)} = 2x^2 + 3x.$  Squaring both sides, we get
\[(1 - x^2)(1 - 4x^2) = (2x^2 + 3x)^2.\]This simplifies to $12x^3 + 14x^2 - 1 = 0.$  Thus, the smallest possible value of $|a| + |b| + |c| + |d|$ is $12 + 14 + 0 + 1 = \boxed{27}.$