Question

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

Answer 1

Answer: 6

$\frac{\sin 3x-\sin 2x}{\sin x}=\frac{3\sin x-4{\sin }^{3}x-2\sin x\cos x}{\sin x}$
$=3-4{\sin }^{2}x-2\cos x$
$=3-4\left(1-{\cos }^{2}x\right)-2\cos x$
$=4{\cos }^{2}x-2\cos x-1$
$={\left(2\cos x-\frac{1}{2}\right)}^2-\frac{5}{4}$
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 $6$ .