Question

Find the principal and general solutions of $\text{cosec}x=-2$

Answer 1

Step $1:$ Solve for principal solution.
Given $\text{cosec}x=-2\Rightarrow \frac{1}{\sin x}=-2\Rightarrow \sin x=-\frac{1}{2}$

We know that $\sin {30}^{\circ }=\frac{1}{2}$
Since $\sin x$ is negative.
$x$ will be in $3^{rd}$ and $4^{th}$ quadrant.

Value in $3^{rd}$ quadrant $={180}^{\circ }+{30}^{\circ }={210}^{\circ }$
Value in $4^{th}$ quadrant $={360}^{\circ }-{30}^{\circ }={330}^{\circ }$
So, principal solutions are $x={210}^{\circ }=210\times \frac{\pi }{180}=\frac{7\pi }{6}$ and $x={330}^{\circ }=330\times \frac{\pi }{180}=\frac{11\pi }{6}$

Step $2:$ Solve for general solution
$\sin x=-\frac{1}{2}$
$\sin x=\sin \frac{7\pi }{6}$

We know that if $\sin x=\sin y$
General solutions is $x=n\pi +(-1{)}^{n}y,$ where $n\in Z$
Put $y=\frac{7\pi }{6}$
Hence, $x=n\pi +(-1{)}^n\frac{7\pi }{6}$ where $n\in Z$