Trigonometric equation with sine graph

Question

If sin^-1(1/2)=30 degrees ,solve the equation I've never seen anything like this and I can't find anything in my textbook so an explanation would be appreciated

Answer 1

If you know identity $\sin (-x)=-\sin x$, and can solve $\sin x=\frac{1}{2}$ then $-x$ will be the solution of your original equation. Please be aware that you'll have two solutions.

If you are given a graph of $y=\sin x$ then you can draw line $y=-\frac{1}{2}$ and your solutions will be points of intersection of the sine graph and that line.

Answer 2

In general : If $\sin x=\sin\alpha,$ then $x=n\pi+(-1)^{n}\alpha$ for any $n\in\Bbb{Z}.$

For your problem, we have $\sin x=\sin (-30^0)$ and choose values of $n$ so that $x$ is in the given range.