Question

The number of solution(s) of the equation $3\tan (x-\frac{\pi }{12})=\tan (x+\frac{\pi }{12})$ in $A=\{x\in R:x^2-6x\le 0\}$ is

Answer 1

$x^2-6x\le 0$
$\Rightarrow x(x-6)\le 0$
$\Rightarrow x\in [0,6]$

$3\tan (x-\frac{\pi }{12})=\tan (x+\frac{\pi }{12})$
$\Rightarrow 3\sin (x-\frac{\pi }{12})\cos (x+\frac{\pi }{12})=\sin (x+\frac{\pi }{12})\cos (x-\frac{\pi }{12})\Rightarrow 3(\sin 2x-\sin \frac{\pi }{6})=\sin 2x+\sin \frac{\pi }{6}\Rightarrow 2\sin 2x=2$
$\Rightarrow \sin 2x=1$
$\Rightarrow 2x=2n\pi +\frac{\pi }{2},n\in Z\Rightarrow x=n\pi +\frac{\pi }{4},n\in Zn=0;x=\frac{\pi }{4}\in An=1;x=\pi +\frac{\pi }{4}=\frac{5\pi }{4}\in An=3;x=2\pi +\frac{\pi }{4}\notin A$
Hence, two solutions are there.