Question

Total number of solutions of equation $\sin x.\tan 4x=\cos x$ belonging to $(-\pi ,2\pi )$ are

• A. $4$
• B. $7$
• C. $8$
• D. $12$

Answer 1

The correct option is D $12$

We have $\sin x\cdot \tan 4x=\cos x$

$\Rightarrow \frac{\sin x}{\cos x}\cdot \tan 4x=1$

$\Rightarrow \tan x\cdot \tan 4x=1$

$\Rightarrow \tan x\cdot \tan 2\left(2x\right)=1$

$\Rightarrow \frac{2\tan 2x}{1-{\tan }^{2}2x}=\frac{1}{\tan x}$

$\Rightarrow \frac{2\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}{1-{\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}^2}=\frac{1}{\tan x}$

$\Rightarrow \frac{2\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}{\frac{(1-{\tan }^{2}x{)}^2-4{\tan }^{2}x}{(1-{\tan }^{2}x{)}^2}}=\frac{1}{\tan x}$

$\Rightarrow \frac{4\tan x(1-{\tan }^{2}x)}{1-2{\tan }^{2}x+{\tan }^{4}x-4{\tan }^{2}x}=\frac{1}{\tan x}$

$\Rightarrow \frac{4\tan x-4{\tan }^{3}x}{1-6{\tan }^{2}x+{\tan }^{4}x}=\frac{1}{\tan x}$

$\Rightarrow 4{\tan }^{2}x-4{\tan }^{4}x=1-6{\tan }^{2}x+{\tan }^{4}x$

$\Rightarrow 5{\tan }^{4}x-10{\tan }^{2}x+1=0$

So, ${\tan }^{2}x=\frac{10\pm \sqrt{100-20}}{10}=\frac{10\pm 4\sqrt{5}}{10}$

$\therefore {\tan }^{2}x=\frac{5-2\sqrt{5}}{5}$ or ${\tan }^{2}x=\frac{5+2\sqrt{5}}{5}$

$\therefore \tan x=\pm {\left(\frac{5-2\sqrt{5}}{5}\right)}^{1/2}$ or $\tan x=\pm {\left(\frac{5+2\sqrt{5}}{5}\right)}^{1/2}$

As $x\in (-\pi ,2\pi )$, $\tan x$ has two positive and two negative values.

Also $\tan x>0\forall x\in (-\pi ,-\frac{\pi }{2})\cup (0,\frac{\pi }{2})\cup (\pi ,\frac{3\pi }{2})$

and $\tan x<0\forall x\in (-\frac{\pi }{2},0)\cup (\frac{\pi }{2},\pi )\cup (\frac{3\pi }{2},2\pi )$

So the number of values of $x$ which satisfy the given equation is $12$.