Question

The general solution of $x-5\sin x\cos x-6{\cos }^{2}x=0$ is

• A. $x=n\pi -\frac{\pi }{4},n\in Z$ only
• B. $n\pi +{\tan }^{-1}\left(6\right),n\in Z$ only
• C. Both (A) and (B)
• D. None of these

Answer 1

The correct option is B $n\pi +{\tan }^{-1}\left(6\right),n\in Z$ only

Dividing the given equation by ${\cos }^{2}x,$ as $\cos x=0$ does not satisfying the equation,

We have ${\tan }^{2}x-5\tan x-6=0\Rightarrow (\tan x+1)(\tan x-1)=0$

$\Rightarrow \tan x=-1$ or $\tan x=6$

If $\tan x=-1=\tan (-\frac{\pi }{4}),$ then $x=n\pi -\frac{\pi }{4},\forall n\in Z$

and if $\tan x=6=\tan \alpha \Rightarrow \alpha =n\pi +{\tan }^{-1}6$,

Hence $x=n\pi -\frac{\pi }{4},n\pi +{\tan }^{-1}6,\forall n\in Z$