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$\begin{aligned} (\tan^2 \phi -2)(\sin^2 \phi - 1) & = 0 \end{aligned}$

For $0 \le \phi \le 2\pi$ ,

$\implies \begin{cases} \tan^2 \phi - 2 = 0 & \implies \tan \phi = \begin{cases} \sqrt 2 & \implies \phi = \begin{cases} \tan^{-1} \sqrt 2 & \small \text{in the first quadrant} \\ \pi + \tan^{-1} \sqrt 2 & \small \text{in the third quadrant} \end{cases} \\ - \sqrt 2 & \implies \phi = \begin{cases} \pi - \tan^{-1} \sqrt 2 & \small \text{in the second quadrant} \\ 2 \pi - \tan^{-1} \sqrt 2 & \small \text{in the fourth quadrant} \end{cases} \end{cases} \\ \sin^2 \phi - 1 = 0 & \implies \sin \phi = \begin{cases} 1 & \implies \phi = \dfrac \pi 2 \\ -1 & \implies \phi = \dfrac {3\pi}2 \end{cases} \end{cases}$

but at $\phi = \dfrac{\pi}{2}$ and at $\phi = \dfrac{3\pi}{2}$ the $\tan$ becomes undefined, and thus whole equation takes the indeterminate form of $\infty \times 0$ , and hence cannot be equal to zero.

Therefore there are 4 solutions for $\phi \in [0, 2\pi]$ and $\boxed{8}$ solutions for $\phi \in [-2\pi, 2\pi]$ .