Question

The number of solutions of the equation, $\frac{\tan 3x}{\tan x}=2$ is

• A. $0$
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is A $0$

$\frac{\tan 3x}{\tan x}=2$
$\tan 3x=2\tan x$
$\sin 3x\cos x=2\sin x\cos 3x$
$\sin 3x\cos x-\sin xcos3x=\sin x\cos 3x$
$\sin (3x-x)=\sin x\cos 3x$ ($\sin (A-B)=\sin A\cos B-\cos A\sin B$)
$\sin 2x=\sin x\cos 3x$
$2\sin x\cos x=\sin x\cos 3x$ ($\sin 2A=2\sin A\cos A$)
$2\cos x=\cos 3x$
$2\cos x=4{\cos }^{3}x-3\cos x$ ($\cos 3A={\cos }^{3}A-3\cos A$)
$4{\cos }^{3}x-5\cos x=0$
$\cos x(4{\cos }^{2}x-5)=0$
$\cos x=0$ or ${\cos }^{2}x=\frac{5}{4}$, ($\cos x=\pm \frac{\sqrt{5}}{2}$ is not possible)
Also, $\cos x=0$ is not possible. Else, $\tan x=\infty$, denominator of the given equation will not have a value.
Thus, the equation has no solution.