Question

Find the magnitude of minimum value of $1+2\sin x+3{\cos }^{2}x$

Answer 1

$1+2\sin x+3{\cos }^{2}x$
$=-3{\sin }^{2}x+2\sin x+4$
$=-3\left({\sin }^{2}x\frac{2\sin x}{3}\right)+4$
$=-3{(\sin x-\frac{1}{3})}^2+\frac{13}{3}$
Now $0\le {(\sin -\frac{1}{3})}^2\le \frac{16}{9}$
$\Rightarrow -\frac{16}{3}\le -3{(\sin x-\frac{1}{3})}^2\le 0$
Adding $\frac{13}{3}$ to the equation,
$-1\le -3{(\sin x-\frac{1}{3})}^2+\frac{13}{3}\le \frac{13}{3}$