Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\sin \left(6x\right) \; \cos x = - \cos \left(6x\right) \; \sin x$
$\sin \left(6x\right) \; \cos x = - \cos \left(6x\right) \; \sin x$
i.e. $\;$ $\sin \left(6x\right) \; \cos x + \cos \left(6x\right) \; \sin x = 0$
i.e. $\;$ $\sin \left(6x + x\right) = 0$
i.e. $\;$ $\sin \left(7 x\right) = 0$
$\implies$ $7 x = n \pi, \;\;\; n \in Z$
i.e. $\;$ $x = \dfrac{n \; \pi}{7}, \;\;\; n \in Z$
In the range $\left[0, 2 \pi \right)$,
when $\;$ $n = 0$, $\;$ $x = 0$
when $\;$ $n = 1$, $\;$ $x = \dfrac{\pi}{7}$
when $\;$ $n = 2$, $\;$ $x = \dfrac{2 \pi}{7}$
when $\;$ $n = 3$, $\;$ $x = \dfrac{3 \pi}{7}$
when $\;$ $n = 4$, $\;$ $x = \dfrac{4 \pi}{7}$
when $\;$ $n = 5$, $\;$ $x = \dfrac{5 \pi}{7}$
when $\;$ $n = 6$, $\;$ $x = \dfrac{6 \pi}{7}$
when $\;$ $n = 7$, $\;$ $x = \pi$
when $\;$ $n = 8$, $\;$ $x = \dfrac{8 \pi}{7}$
when $\;$ $n = 9$, $\;$ $x = \dfrac{9 \pi}{7}$
when $\;$ $n = 10$, $\;$ $x = \dfrac{10 \pi}{7}$
when $\;$ $n = 11$, $\;$ $x = \dfrac{11 \pi}{7}$
when $\;$ $n = 12$, $\;$ $x = \dfrac{12 \pi}{7}$
when $\;$ $n = 13$, $\;$ $x = \dfrac{13 \pi}{7}$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = 0, \; \dfrac{\pi}{7}, \; \dfrac{2\pi}{7}, \; \dfrac{3\pi}{7}, \cdots, \pi, \; \cdots, \dfrac{13 \pi}{7}$