Find the principal and general solution of the following equation $cosec x=-2$

$\begin{array}{1 1}(A)\;\large\frac{7\pi}{6},\frac{11\pi}{3},\normalsize x=n\pi+(-1)^n\large\frac{7\pi}{6}\\(B)\;\large\frac{7\pi}{3},\frac{2\pi}{3},\normalsize x=n\pi+(-1)^n\large\frac{\pi}{2}\\(C)\;\large\frac{\pi}{5},\frac{4\pi}{5},\normalsize x=n\pi+(-1)^n\large\frac{\pi}{5}\\(D)\;\large\frac{\pi}{7},\frac{4\pi}{7},\normalsize x=n\pi+(-1)^n\large\frac{\pi}{7}\end{array}$

Toolbox:
• $cosec \large\frac{\pi}{6}$$=2,\sin \large\frac{\pi}{6}=\frac{1}{2} • \sin x=\sin y • \therefore x=n\pi+(-1)^ny cosec (\pi+\large\frac{\pi}{6})$$=-cosec \large\frac{\pi}{6}=$$-2 \sin (\pi+\large\frac{\pi}{6})$$=-\sin \large\frac{\pi}{6}=-\frac{1}{2}$
$cosec \large\frac{7\pi}{6}=$$-2 \sin (2\pi-\large\frac{\pi}{6})=$$-\sin \large\frac{\pi}{6}=-\frac{1}{2}$
$\Rightarrow \sin \large\frac{11}{6}$$\pi=-\large\frac{1}{2} cosec \large\frac{11\pi}{6}$$=-2$
Principal solutions are $\large\frac{7\pi}{6},\frac{11\pi}{6}$
$\therefore cosec x=cosec \large\frac{7\pi}{6}$
General solution $x=n\pi+(-1)^n\large\frac{7\pi}{6}$ where $n \in z$
Hence (A) is the correct answer.