Question

How do you solve `cos2x=sin^2x-2`?

Answer 1

`pi/2 + 2kpi`
`(3pi)/2 +2kpi`

Explanation:

Use identity:
cos 2x = 1 - 2sin^2 x
In this case:
1 - 2sin^2 x = sin^2 x - 2
`3sin^2 x = 3` -- >`sin ^2 x = 1`--> `sin x = +- 1`
Unit circle -->
sin x = 1 -->`x = pi/2 + 2kpi`
sin x = -1 --> `x = (3pi)/2 +2kpi`