Solve the equation: $\;$ $x^{\left(5 \sin 3x\right) + 2} = \dfrac{1}{\sqrt{x}}$, $\;$ for $\;$ $x \in \left[0, 2 \pi\right]$
Given equation: $\;\;$ $x^{\left(5 \sin 3x\right) + 2} = \dfrac{1}{\sqrt{x}}$, $\;$ for $\;$ $x \in \left[0, 2 \pi\right]$
i.e. $\;$ $x^{\left(5 \sin 3x\right) + 2} = x^{\frac{-1}{2}}$
i.e. $\;$ $x^{\left(5 \sin 3x\right) + 2 + \frac{1}{2}} = 1$
i.e. $\;$ $x^{\left(5 \sin 3x\right) + \frac{5}{2}} = x^0$
$\implies$ $5 \sin 3x + \dfrac{5}{2} = 0$
i.e. $\;$ $\sin 3x = \dfrac{-1}{2}$
i.e. $\;$ $3x = 2 n \pi + \pi + \dfrac{\pi}{6}$; $\;$ $3x = 2n \pi + 2 \pi - \dfrac{\pi}{6}$
i.e. $\;$ $3x = 2 n \pi + \dfrac{7 \pi}{6}$; $\;$ $2 n \pi + \dfrac{11 \pi}{6}$
i.e. $\;$ $x = \dfrac{2n \pi}{3} + \dfrac{7 \pi}{18}$; $\;$ $x = \dfrac{2 n \pi}{3} + \dfrac{11 \pi}{18}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{2n \pi}{3} + \dfrac{7 \pi}{18}, \; \dfrac{2 n \pi}{3} + \dfrac{11 \pi}{18} \right\}$, $\;$ $x \in \left[0, 2\pi\right]$