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

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

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

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

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

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

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