Question

Number of integers in the range of $\frac{\sin 3 x-\sin 2 x}{\sin x}$ is

Answer 1

$\frac{\sin 3 x-\sin 2 x}{\sin x}=\frac{3 \sin x-4 \sin ^{3} x-2 \sin x \cos x}{\sin x} $
$=3-4 \sin ^{2} x-2 \cos x$
$=3-4\left(1-\cos ^{2} x\right)-2 \cos x $
$=4 \cos ^{2} x-2 \cos x-1 $
$=\left(2 \cos x-\frac{1}{2}\right)^{2}-\frac{5}{4}$
Range is $\left[-\frac{5}{4}, 5\right)$
$5$ is not include
$\because \cos x=1$ is not possible otherwise $\sin$ $x=0$
$\therefore $ number of integers in the range is $6$ .