Question

How do you solve `sin^2 x = 2cosx + 2`?

Answer 1

`sin^2(x) = 2cos(x)+2`

`1-cos^2(x) = 2cos(x)+2`

`cos^2(x)+2cos(x)+1 = 0`

`(cos(x)+1)^2 =0`

`cos(x) = (-1)`

Assuming we restrict `x` to `[0,2pi)`

`x=pi`