(b)

We have,

$\cos ^4 x+\frac{1}{\cos ^2 x}=\sin ^4 x+\frac{1}{\sin ^2 x}$

$\Rightarrow \quad \cos ^4 x-\sin ^4 x=\frac{1}{\sin ^2 x}-\frac{1}{\cos ^2 x}$

$\Rightarrow \quad\left(\cos ^2 x-\sin ^2 x\right)\left(\cos ^2 x+\sin ^2 x\right)$

$=\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cos ^2 x}$

$\Rightarrow \quad \cos ^2 x-\sin ^2 x=\frac{\cos ^2 x-\sin ^2 x}{\sin ^2 x \cos ^2 x}$

$\Rightarrow \quad \cos 2 x\left(1-\frac{4}{\sin ^2 2 x}\right)=0$

$\Rightarrow \quad \cos 2 x=0$ or $\sin ^2 2 x=4$

$\therefore \quad \cos 2 x=0$ or $\sin ^2 2 x \neq 4$

$\Rightarrow \quad 2 x=(2 n+1) \frac{\pi}{2}$

$\Rightarrow \quad x=(2 n+1) \frac{\pi}{4}$

$\ln x \in(0,2 \pi)$

$x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$

$\therefore$ Total number of solution $=4$