Question

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

• A. Principal solution $=\frac{\pi }{6},\frac{11\pi }{6}$ and General solution $=n\pi +(-1{)}^n\frac{\pi }{6},n\in Z$
• B. Principal solution $=\frac{7\pi }{2},\frac{11\pi }{2}$ and General solution $=n\pi +\frac{7\pi }{2},n\in Z$
• C. Principal solution $=\frac{7\pi }{6},\frac{11\pi }{6}$ and General solution $=n\pi +(-1{)}^n\frac{7\pi }{6},n\in Z$
• D. Principal solution $=\frac{\pi }{2},\frac{11\pi }{2}$ and General solution $=n\pi +\frac{7\pi }{6},n\in Z$

Answer 1

The correct option is C Principal solution $=\frac{7\pi }{6},\frac{11\pi }{6}$ and General solution $=n\pi +(-1{)}^n\frac{7\pi }{6},n\in Z$

Given: $\text{cosec}x=-2$

$\Rightarrow \frac{1}{\sin x}=-2$

$\Rightarrow \sin x=-\frac{1}{2}$

we know that ${\sin 30}^o=-\frac{1}{2}$

But we need negative value of $\sin x$, $\sin$ is negative in $III$ & $IV$ quadrant.

Value in $III$ quadrant $={180}^o+{30}^o={210}^o$

Value in $IV$ quadrant $={360}^o-{30}^o={330}^o$

so principal solutions are

$x={210}^o=\frac{7\pi }{6}$

$x={330}^o={330}^o\times \frac{\pi }{180}=\frac{11\pi }{6}$

General solution:

Let $\sin x=\sin y$ ..... $\left(1\right)$

$\sin x=-\frac{1}{2}$ ......... $\left(2\right)$

from $\left(1\right)$ & $\left(2\right)$

$\sin y=-1/2$

$\sin y=\sin \frac{7\pi }{6}$

$\Rightarrow y=\frac{7\pi }{6}$

$\therefore \sin x=\sin y$

General solution is

$x=n\pi +{(-1)}^{n}y,n\in I$

$x=n\pi +{(-1)}^n\left(\frac{7\pi }{6}\right),n\in I$