A) \[2n\pi +\frac{\pi }{4},n\in Z\]
B) \[2n\pi -\frac{\pi }{4},n\in Z\]
C) \[n\pi -\frac{\pi }{4},n\in Z\]
D) \[n\pi +\frac{\pi }{4},n\in Z\]
Correct Answer: D
Solution :
\[1+{{\sin }^{2}}x=3\sin x.cosx,tanx\ne \frac{1}{2}\] Divided by \[{{\cos }^{2}}x\] on both sides, \[\frac{1}{{{\cos }^{2}}x}+\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}=3\frac{\sin x.\cos x}{\cos x.\cos x}\] \[{{\sec }^{2}}x+{{\tan }^{2}}x=3\tan x\] \[1+{{\tan }^{2}}x+{{\tan }^{2}}x=3\tan x\] \[2{{\tan }^{2}}x-3\tan x+1=0\] \[2{{\tan }^{2}}x-2\tan x-\tan x+1=0\] \[2\tan x(\tan x-1)-1(\tan x-1)=0\] \[(\tan x-1)(2\tan x-1)=0\] \[\tan x=1,\frac{1}{2}\] We take, \[\tan \,x=1\] \[\left( \because \,\,\tan x\ne \frac{1}{2} \right)\] \[\tan x=\tan (\pi /4)\] \[x=n\pi +\pi /4,n\in Z\]
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