Solve for the value of x in the interval [0,2pi)

Question

Solve for the value of x in the interval [0,2pi)

1) -6 sin^2 x = 3 cos x -3

Answer 1

-6sin²x - 3cosx + 3 = 0

Divide by -3 to get 2sin²x + cosx -1 = 0

Since cos²x + sin²x = 1, sin²x = 1 - cos²x.

So, 2(1 - cos²x) + cosx - 1 = 0

-2cos²x + cosx + 1 = 0

2cos²x - cosx - 1 = 0

(cosx - 1)(2cosx + 1) = 0

cosx = 1 or cosx = -1/2

If cosx = 1, then x = 0

If cosx = -1/2, then x = 2π/3 or 4π/3