A) 0
B) 1
C) 2
D) 3
Correct Answer: C
Solution :
Since, \[1-cos\text{ }x\ge 0,\text{ }sin\text{ }x\ge 0\] Squaring, given equation, we get \[1-\cos x={{\sin }^{2}}x\] \[\Rightarrow \] \[{{\cos }^{2}}x-\cos x=0\] \[\Rightarrow \] \[\cos x=0\]or\[\cos x=1\] If\[cos\text{ }x=0,\]then \[x=2n\pi +\frac{\pi }{2},n\in Z\] If \[cos\text{ }x=1,\]then \[x=2\pi k,k\in Z\] But\[sin\text{ }x\ge 0\]and\[x\in (\pi ,3\pi ),\]we get \[x=2\pi \]or\[x=\frac{5\pi }{2}\]You need to login to perform this action.
You will be redirected in
3 sec