Question

If $p\in Z,$ then the number of solutions (in $x$) of the equation $\left(\frac{p}{2p-3}\right){\sin }^{2}x+\left(\frac{2p-3}{p}\right){\text{cosec}}^{2}x=2$ in $x\in [0,2\pi ]$ is

Answer 1

Let $\left(\frac{p}{2p-3}\right){\sin }^{2}x=t$
Then, $t+\frac{1}{t}=2$
$\Rightarrow t^2-2t+1=0\Rightarrow (t-1{)}^2=0\Rightarrow t=1$

$\therefore \left(\frac{p}{2p-3}\right){\sin }^{2}x=1\Rightarrow {\sin }^{2}x=\frac{2p-3}{p}$
As $0\le \frac{2p-3}{p}\le 1$
$\Rightarrow p\in [\frac{3}{2},3]$
Since $p\in Z,$
$\therefore p=2,3$

For $p=2,$
${\sin }^{2}x=\frac{1}{2}={\sin }^2\frac{\pi }{4}\Rightarrow x=n\pi \pm \frac{\pi }{4}\Rightarrow x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

For $p=3,$
${\sin }^{2}x=1\Rightarrow \sin x=\pm 1\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2}$

Hence, number of values of $x\in [0,2\pi ]$ is $6$