<s>[INST] Solve $\sin^2 x − \cos^2 x = \sin x$, when $x\in -\pi ≤ x ≤\pi$ I have to solve for $x$ using the domain of $-\pi ≤ x ≤\pi$
$$\sin^2 x − \cos^2 x = \sin x $$
I tried changing $\cos^2 x$  to $1 - \sin^2 x$, and then getting
$$\sin^2 x - 1 + \sin^2 x = \sin x \to 2\sin^2 x - 1 = \sin x$$ 
Then I have no clue where to go from there. Please help!
 [/INST] From
$\sin^2 x − \cos^2 x = \sin x$,
since
$\cos^2 x = 1-\sin^2 x$,
$\sin x 
= \sin^2 x - (1-\sin^2 x)
=2\sin^2 x -1
$.
Now we only have $\sin x$,
so let $\sin x = y$.
This becomes
$y = 2y^2-1$.
You can solve this for $y$.
From the possible values of $y$,
you can then get $x$.
I will leave it at that.
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