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), \[{{\sin }^{2}}2x=2{{\cos }^{2}}x\] Þ\[2{{\cos }^{2}}x\cos 2x=0\]Þ\[x=(2n+1)\frac{\pi }{2}\text{ or }x=(2n+1)\frac{\pi }{4}\] \[\therefore \] Common roots are \[(2n\pm 1)\frac{\pi }{4}\] Solving (ii), \[\frac{2\tan x+1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\tan x\] \[\Rightarrow \] \[{{\tan }^{3}}x+{{\tan }^{2}}x-\tan x-1=0\] \[\Rightarrow \] \[({{\tan }^{2}}x-1)\,(\tan x+1)=0\] \[\Rightarrow \] \[x=m\pi \pm \frac{\pi }{4}\] Trick: For\[n=0\], option gives \[\theta =-\frac{\pi }{2}\] which satisfies the equation (i) but does not satisfy the (ii). Now option gives \[\theta =\frac{\pi }{4}\] which satisfies both the equations.
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