Solve the equation: $\;$ $\sqrt{2} \cos^2 \left(7x\right) - \cos \left(7x\right) = 0$
Given equation: $\;\;\;$ $\sqrt{2} \cos^2 \left(7x\right) - \cos \left(7x\right) = 0$
i.e. $\;$ $\cos \left(7x\right) \left[\sqrt{2} \cos \left(7x\right) - 1\right] = 0$
$\implies$ $\cos \left(7x\right) = 0$ $\;\;$ or $\;\;$ $\sqrt{2} \cos \left(7x\right) - 1 = 0$
When $\;\;$ $\cos \left(7x\right) = 0$
$\implies$ $7x = \dfrac{\left(2n + 1\right) \pi}{2}, \;\;\; n \in I$
i.e. $\;$ $x = \dfrac{\pi}{14} \left(2n + 1\right), \;\;\; n \in I$
When $\;\;$ $\sqrt{2} \cos \left(7x\right) - 1 = 0$
i.e. $\;$ $\cos \left(7x\right) = \dfrac{1}{\sqrt{2}}$
i.e. $\;$ $7x = \pm \cos^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + 2 m \pi, \;\;\; m \in I$
i.e. $\;$ $7x = \pm \dfrac{\pi}{4} + 2 m \pi, \;\;\; m \in I$
i.e. $\;$ $x = \pm \dfrac{\pi}{28} + \dfrac{2}{7} m \pi, \;\;\; m \in I$
i.e. $\;$ $x = \dfrac{\pi}{28} \left(8m \pm 1\right), \;\;\; m \in I$