Question

Find the general value of x for which $\sqrt{3}$ cosec x =2.

Answer 1

Given: $\sqrt{3}cosecx=2\Rightarrow cosecx=\frac{2}{\sqrt{3}}\Rightarrow sinx=\frac{\sqrt{3}}{2}.$

The least value of $xin[0,2\pi )$ [ $\text{for which}$ $sinx=\frac{\sqrt{3}}{2}$ $is$ $\frac{\pi }{3}.$

$\therefore sinx=sin\frac{\pi }{3}\Rightarrow x=\{n\pi +(-1{)}^n.\frac{\pi }{3}\}$, where $n\in I.$

Hence, the general solution is $x=\{n\pi +(-1{)}^n.\frac{\pi }{3}\}$, where $n\in I.$