Recall the triple angle identity for sine:
sin(3x) = 3 sin(x) cos²(x) - sin³(x)
… = 3 sin(x) (1 - sin²(x)) - sin³(x)
… = 3 sin(x) - 4 sin³(x)
So the equation is equivalent to
sin²(x) - 2 sin(x) + 2 = sin²(3x)
sin²(x) - 2 sin(x) + 2 = (3 sin(x) - 4 sin³(x))²
sin²(x) - 2 sin(x) + 2 = 9 sin²(x) - 24 sin⁴(x) + 16 sin⁶(x)
16 sin⁶(x) - 24 sin⁴(x) + 8 sin²(x) + 2 sin(x) - 2 = 0
or, by replacing y = sin(x),
16y ⁶ - 24y ⁴ + 8y ² + 2y - 2 = 0
By inspection, y = 1 is a root, since
16 - 24 + 8 + 2 - 2 = 0
so we get one family of solutions,
sin(x) = 1 ===> x = arcsin(1) + 2nπ = π/2 + 2nπ
Now suppose y ≠ 1. Dividing both sides of the sextic equation by y - 1 then gives
(16y ⁶ - 24y ⁴ + 8y ² + 2y - 2)/(y - 1) = 0
16y ⁵ + 16y ⁴ - 8y ³ - 8y ² + 2 = 0
Unfortunately, the remaining roots can only be obtained through numerical methods. Using a calculator, you would find
y ≈ -1.127
y ≈ -0.475 ± 0.368i
y ≈ 0.539 ± 0.127i
But if you're solving over the reals, you can ignore these solutions. Remember that sin(x) is bounded between -1 and 1, and its value goes outside this interval only when x is complex.
