Consider the equation lo g _ 2 x (1 + x) = 2, x ∈ [- (π)/(2),...

$\log_{\sqrt{2}\sin x}\left(1 + \cos x\right) = 2$ ….(i)
$\sqrt{2}\sin x \ne 1, \sqrt{2}\sin x > 0, 1 + \cos x > 0$
$⇒ \sin x \ne \dfrac{1}{\sqrt{2}}, \sin x > 0$ and $x \ne 0$ or a multiple of $\pi$
$⇒ x \in \left(0, \pi\right) - \left\{\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right\}$ (feasible region)

Hence, from (i), we get,

${\left(\sqrt{2}\sin x\right)}^{2} = 1 + \cos x$
$⇒ 2si{n}^{2}x = 1 + \cos x$
$⇒ 2co{s}^{2}x + \cos x - 1 = 0$
$⇒ \left(2\cos x - 1\right)\left(\cos x + 1\right) = 0$
$⇒ \cos x = \dfrac{1}{2}$ $\ldots \left[\cos x + 1 > 0\right]$
$⇒ x = \dfrac{\pi}{3}$
$⇒ p = 1, q = 3$
$⇒ {p}^{2} + {q}^{2} = 10$