Trigonometry

Rearrange the terms to get

$\cos^{4}x - \sin^{4}x = \dfrac{1}{\sin^{2}x} - \dfrac{1}{\cos^{2}x}$

$\Longrightarrow (cos^{2}x - \sin^{2}x)(\cos^{2}x + \sin^{2}x) = \dfrac{\cos^{2}x - \sin^{2}x}{\sin^{2}x * \cos^{2}x}$

$\Longrightarrow (\cos^{2}x - \sin^{2}x)(1 - \frac{1}{\sin^{2}x * \cos^{2}x}) = 0$ .

So either $\cos^{2}x = \sin^{2}x \Longrightarrow \cos x = \pm \sin x$ , which has $4$ solutions on the given interval, namely $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}$ and $\frac{7\pi}{4}$ , or $\sin^{2}x * \cos^{2}x = 1 \Longrightarrow \sin^{2}(2x) = 4$ , which has no solutions, since the range of the sine function is $[-1, 1]$ .

Thus there are $\boxed{4}$ solution on the given interval.