How do you go about solving this question?

$\dfrac{1}{2}\sec\left(2\phi+\dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt3}$

$\sec\left(2\phi+\dfrac{\pi}{4}\right) = \dfrac{2}{\sqrt3}$

$\cos\left(2\phi+\dfrac{\pi}{4}\right) = \dfrac{\sqrt3}{2}$

This means that  $2\phi+\dfrac{\pi}{4} = \dfrac{\pi}{6} + 2k\pi$ or $2\phi+\dfrac{\pi}{4} = -\dfrac{\pi}{6} + 2k\pi$, $k \in \mathbb{Z}$ (remember that cosine is an even periodic function).

which means that $\phi = -\dfrac{\pi}{24} + k\pi$ or $\phi = -\dfrac{5\pi}{24} + k\pi, k \in \mathbb{Z}$ 

Now, since we need $-\pi < \phi < \pi$, by inspection we can see that the solutions must be:

$\phi = -\dfrac{\pi}{24}, -\dfrac{\pi}{24}+ \pi, -\dfrac{5\pi}{24}, -\dfrac{5\pi}{24} + \pi$, which coincides with the given answers.