Question

Number of solution(s) of the equation $\ln (2-{\sin }^{2}x)=0$ where $x\in [0,10\pi ]$ is

Answer 1

$\ln (2-{\sin }^{2}x)=0$
$\Rightarrow 2-{\sin }^{2}x=1$
$\Rightarrow {\sin }^{2}x=1={\sin }^2\frac{\pi }{2}$
$\Rightarrow x=n\pi \pm \frac{\pi }{2}$
Now, $x\in [0,10\pi ]$
$\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\cdots ,\frac{19\pi }{2}$
Number of solutions $=10$