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 :
| \[|\cos x|\,\,=\cos x-2\sin x\] |
| Case I when \[\cos x\ge 0\] |
| then \[\cos x=\cos x-2\sin x\] |
| \[\Rightarrow \] \[\sin x=0\] \[\Rightarrow \] \[x=n\pi \] |
| But \[\cos x>0\] \[\Rightarrow \] \[\cos x=1,\]\[x=2\,m\pi \] |
| 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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