${(3{\cos ^2}x + 2\sin x + 1)_{\max }} = ?$

The given expression = $- 3{\sin ^2}x + 2\sin x + 4$

= – $\left[ {{{(\sqrt 3 \sin x)}^2} – 2.\frac{1}{{\sqrt 3 }}.\sqrt 3 \sin x + \frac{1}{3} – \frac{1}{3}} \right]$ + 4

$ = – \left[ {{{\left( {\sqrt 3 \sin x – \frac{1}{{\sqrt 3 }}} \right)}^2} – \frac{1}{3}} \right] + 4$

$ = 4 + \frac{1}{3} – {\left( {\sqrt 3 \sin x – \frac{1}{{\sqrt 3 }}} \right)^2}$

$ = \frac{{13}}{3} – (non-negative\,number)$

$\therefore {(3{\cos ^2}x + 2\sin x + 1)_{\max }} = \frac{{13}}{3}$