(c) $3{\sin ^2}x – 7\sin x + 2 = 0$

$ \Rightarrow $ $3{\sin ^2}x – 6\sin x – \sin x + 2 = 0$

$ \Rightarrow $ $3\sin (\sin x – 2) – (\sin x – 2) = 0$

$ \Rightarrow $ $(3\sin x – 1)\,(\sin x – 2) = 0$

$ \Rightarrow $ $\sin x = \frac{1}{3}{\rm{ or 2}}$

$ \Rightarrow $ $\sin x = \frac{1}{3}$, ($ \because \,\,\sin x \ne 2$)

Let ${\sin ^{ – 1}}\frac{1}{3} = \alpha $,

$0 < \alpha < \frac{\pi }{2}$ are the solutions in $[0,{\rm{ }}5\pi ]$. 

Then $\alpha ,$$\pi – \alpha ,\,$$2\pi + \alpha ,$ $\,3\pi – \alpha ,$ $\,4\pi + \alpha $, $5\pi – \alpha $ are the solutions in $[0,\,5\pi ]$.

$\therefore $ Required number of solutions $= 6$.