Solve $\sin^2 x-1=0$ for $0^\circ\le x\le360^\circ$

Question

I am new to this so I don't know how to type the exponent. Basically, I have to solve for $x$ for the following equation

$$\sin^2 x-1=0.$$

Answer 1

Hint: Let's say that $y=\sin(x)$. What does the value of $y$ have to be in order to satisfy your equation?

Answer 2

HINT: $$\sin^2x-1=(\sin x-1)(\sin x+1).$$

Answer 3

$$\sin^2x=1\iff\cos2x=1-2\sin^2x=-1=\cos180^\circ$$

$$\iff2x=(2n+1)180^\circ\iff x=(2n+1)90^\circ$$ where $n$ is any integer

Find suitable values of $n$ from the given condition