cosx=-sin^2 x-1 find all solution in interval [0,2pi)

Question

please help me find the solutions

Answer 1

cos²x + sin²x = 1, so sin²x = 1 - cos²x

The given equation is therefore equivalent to cosx = -(1 - cos²x) - 1

So, cos²x - cosx - 2 = 0

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

cosx = 2 or cosx = -1

The equation cosx = 2 has no solution since -1 ≤ cosx ≤ 1

If cosx = -1 and x is in the interval [0, 2π), x = π is the only solution.