Question

How do you solve `cos^2 x-sin^2 x=sin x` over the interval `-pi<x<=pi`?

Answer 1

Write: `cos^2(x)=1-sin^2(x)` so that you get:

`1-sin^2(x)-sin^2(x)-sin(x)=0`
`-2sin^2(x)-sin(x)+1=0`

Set: `sin(x)=t` and get:

`-2t^2-t+1=0`
`t_1=-1`
`t_2=1/2`

but `sin(x)=t`

And in your interval:

`sin(x)=-1` and `x=-pi/2`
`sin(x)=1/2` and `x=pi/6 and x=5pi/6`

Hope it helps.