Solve $\sin^2 x − \cos^2 x = \sin x$, when $x\in -\pi ≤ x ≤\pi$

Question

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!

Answer 1

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.