Question

Find range
$\frac{1}{8-3\sin x}$

Answer 1

We know that $-1\le \sin x\le 1$

$\Rightarrow -3\le 3\sin x\le 3$

$\Rightarrow 3\ge -3\sin x\ge -3$ (multiplying by $-3$)

$\Rightarrow 8+3\ge 8-3\sin x\ge 8-3$

$\Rightarrow 11\ge 8-3\sin x\ge 5$

$\Rightarrow \frac{1}{11}\le \frac{1}{8-3\sin x}\le \frac{1}{5}$

$\therefore$ Range is $[\frac{1}{11},\frac{1}{5}]$