Question

Find all the solutions in the interval [0,2π)
sin²x = 2-2cosX

Answer 1

Explanation:

sin²x = 2 - 2cosx

1 - cos²x = 2 - 2cosx (because sin²x + cos²x = 1)

cos²x - 2cosx + 1 = 0

(cosx - 1)² = 0

cosx - 1 = 0

cosx = 1

Hence x = 0.