A) \[x=n\pi \]
B) \[x=n\pi +\frac{\pi }{4}\]
C) \[x=n\pi +{{(-\,1)}^{n}}\frac{\pi }{4}\]
D) \[x=(2n+1)\,\pi +\frac{\pi }{4},\] \[n\in I\]
Correct Answer: D
Solution :
| \[\left| \cos x \right|=\cos x-2\sin x\] |
| Case I when \[\cos x\ge 0\]then \[\cos x=\cos x-2\sin x\]\[\Rightarrow \sin x=0\]\[\Rightarrow x=n\pi \]But \[\cos x>0\]\[\Rightarrow \cos x=1,x=2m\pi \] |
| Case II when \[\cos x<0\]then \[-\,\cos x=\cos x-2\sin x\] |
| \[\cos x=\sin x\]\[\Rightarrow \tan x=1\]\[\Rightarrow \tan x=1,\]\[\cos x<0\]\[\Rightarrow x=(2n+1)\,\pi +\frac{\pi }{4},\,\,n\in I\] |
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