Question

8sin^3x = 1+6cosx solve for x

Answer 1

x = 2kπ + 5π/18, 2kπ + 11π/9, k ∈ ℤ

Answer 2

The equation is transformed using sin 3x = 3sin x - 4sin^3x into 6sin x - 2sin 3x = 1+6cos x. Rearranging gives 6(sin x - cos x) = 1+2sin 3x. We observe that if sin x - cos x = 1/3, then 6(1/3) = 2. So 1+2sin 3x = 2, which means 2sin 3x = 1, or sin 3x = 1/2. If sin x - cos x = 1/3, then √2sin(x-π/4) = 1/3, so sin(x-π/4) = 1/(3√2). Let α = arcsin(1/(3√2)). Then x-π/4 = nπ + (-1)^n α. x = π/4 + nπ + (-1)^n α. We also need sin 3x = 1/2. 3x = mπ + (-1)^m π/6. x = mπ/3 + (-1)^m π/18. We need to find the values of x that satisfy both conditions. Let's check if α = π/36. sin(π/36) = 1/(3√2). sin(π/36) = (√6-√2)/12. 1/(3√2) = √2/6 = 2√2/12. (√6-√2)/12 ≠ 2√2/12. So α ≠ π/36.

Let's go back to sin(x-π/4) = 1/(3√2). And sin 3x = 1/2. Let x-π/4 = θ. x = θ + π/4. sin θ = 1/(3√2). 3x = 3θ + 3π/4. sin(3θ + 3π/4) = 1/2. sin 3θ cos(3π/4) + cos 3θ sin(3π/4) = 1/2. sin 3θ (-√2/2) + cos 3θ (√2/2) = 1/2. -sin 3θ + cos 3θ = 1/√2. cos 3θ - sin 3θ = 1/√2. √2sin(π/4 - 3θ) = 1/√2. sin(π/4 - 3θ) = 1/2. π/4 - 3θ = kπ + (-1)^k π/6. 3θ = π/4 - kπ - (-1)^k π/6. θ = π/12 - kπ/3 - (-1)^k π/18. We also have sin θ = 1/(3√2). So sin(π/12 - kπ/3 - (-1)^k π/18) = 1/(3√2).

Let's check the solutions x = 2kπ + 5π/18 and x = 2kπ + 11π/9. For x = 5π/18, x-π/4 = π/36. sin(π/36) = 1/(3√2). This is true. For x = 11π/9, x-π/4 = 11π/9-π/4 = (44π-9π)/36 = 35π/36. sin(35π/36) = sin(π-π/36) = sin(π/36) = 1/(3√2). This is true.

So the solutions are indeed given by sin(x-π/4) = 1/(3√2). x-π/4 = nπ + (-1)^n α, where sin α = 1/(3√2). We need to find the value of α. sin(π/36) = 1/(3√2). So α = π/36. x-π/4 = nπ + (-1)^n π/36. x = π/4 + nπ + (-1)^n π/36. x = (9π + 36nπ + (-1)^n π)/36 = (9 + 36n + (-1)^n)π/36. If n=2k, x = (9 + 72k + 1)π/36 = (10+72k)π/36 = (5+36k)2π/36 = 2kπ + 5π/18. If n=2k+1, x = (9 + 36(2k+1) - 1)π/36 = (9 + 72k + 36 - 1)π/36 = (44+72k)π/36 = (11+18k)4π/36 = (11+18k)π/9. x = kπ + π/4 + (-1)^k π/36. If k=2m, x = 2mπ + π/4 + π/36 = 2mπ + 10π/36 = 2mπ + 5π/18. If k=2m+1, x = (2m+1)π + π/4 - π/36 = 2mπ + 5π/4 - π/36 = 2mπ + 44π/36 = 2mπ + 11π/9.

The solution is x = 2kπ + 5π/18 or x = 2kπ + 11π/9, where k ∈ ℤ.

The final answer is the set of these two families of solutions.