6 sin2 Θ - 11 sin Θ = -4

Question

Find all solutions of each equation in the interval [0,2π).

6 sin² Θ - 11 sin Θ = -4

Please say why some solutions will work and some will not.

Thank you!

Answer 1

6sin²θ - 11sinθ + 4 = 0

Let u = sinθ.

Then, 6u² - 11u + 4 = 0

(3u - 4)(2u - 1)= 0

u = 4/3 or u = 1/2

So, sinθ = 4/3 (impossible since -1 ≤ sinθ ≤ 1) or sinθ = 1/2

θ = SIn^-1(1/2) or π - Sin^-1(1/2) = π/6, 5π/6