Sin^2 x - 2sinx + 2 = sin^2 3x

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.

step-by-step explanation:

(3rd option) - c