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), \[\frac{2\tan x+1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\tan x\]\[\Rightarrow \,\,{{\tan }^{3}}x+{{\tan }^{2}}x-\tan x-1=0\] | |
\[\Rightarrow \,({{\tan }^{2}}x-1)\,(\tan x+1)=0\Rightarrow x=m\pi \pm \pi /4\] |
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