JEE Main & Advanced Sample Paper JEE Main - Mock Test - 2

Common roots of the equations \[2{{\sin }^{2}}x+{{\sin }^{2}}2x=2\] and \[2x+\cos 2x=\tan x,\] are

A) \[x=(2n-1)\frac{\pi }{2}\]

B) \[x=(2n+1)\frac{\pi }{4}\]

C) \[x=(2n+1)\frac{\pi }{3}\]

D) None of these

Correct Answer: B

Solution :

\[2{{\sin }^{2}}x+{{\sin }^{2}}2x=2\]	...(i)
and \[\sin 2x+\cos \,2x=\tan x\]	...(ii)
Solving (i), \[{{\sin }^{2}}2x=2{{\cos }^{2}}x\]
\[\Rightarrow \,\,2co{{s}^{2}}x\,\cos 2x=0\Rightarrow x=(2n+1)\frac{\pi }{2}\] or \[(2n+1)\frac{\pi }{4}\]
\[\therefore \] Common roots are \[(2n\pm 1)\frac{\pi }{4}\]
Solving (ii), \[\frac{2\tan x+1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\tan x\]\[\Rightarrow \,\,{{\tan }^{3}}x+{{\tan }^{2}}x-\tan x-1=0\]
\[\Rightarrow \,({{\tan }^{2}}x-1)\,(\tan x+1)=0\Rightarrow x=m\pi \pm \pi /4\]