What is the correct general solution of $\sin x=-\frac{1}{2}$? My attempt:
$$\sin x=-\frac{1}{2}$$
$$x=n\pi+(-1)^{n}\sin^{-1}(-\frac{1}{2})$$
$$x=n\pi+(-1)^{n}(-\frac{\pi}{6})$$
$$x=n\pi-(-1)^{n}(\frac{\pi}{6})$$
My book's solution:
$$x=n\pi+(-1)^{n}(\frac{7\pi}{6})$$
Who is correct?

We have that
$$\sin x=-\frac{1}{2} \iff x=-\frac \pi 6+2k\pi \quad \lor \quad \frac {7\pi} 6+2k\pi$$
which is equivalent to both solutions indeed

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*for $n=2k$
$$x=2k\pi-(-1)^{2k}\left(\frac{\pi}{6}\right)=2k\pi-\frac{\pi}{6}$$
$$x=2k\pi+(-1)^{2k}\left(\frac{7\pi}{6}\right)=2k\pi+\frac{7\pi}{6}$$

*

*for $n=2k+1$
$$x=(2k+1)\pi-(-1)^{2k+1}\left(\frac{\pi}{6}\right)=2k\pi+\frac{7\pi}{6}$$
$$x=(2k+1)\pi+(-1)^{2k+1}\left(\frac{7\pi}{6}\right)=2k\pi-\frac{\pi}{6}$$
