A) 1
B) 2
C) 3
D) 4
Correct Answer: A
Solution :
\[\left( \cos \frac{x}{2}-2\sin x \right)\sin x+\left( 1+\sin \frac{x}{4}-2\cos x \right)\cos x=0\]\[\Rightarrow \,\,\left( \sin x\,\,\cos \frac{x}{4}+\cos x\sin \frac{x}{4} \right)+\cos x-2\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)=0\]\[\Rightarrow \,\,\sin \left( x+\frac{x}{4} \right)+\cos x-2(1)=0\Rightarrow \sin \frac{5x}{4}+\cos x=2\]\[\Rightarrow \,\,\sin \frac{5x}{4}=\cos x=1\] \[\Rightarrow \,\,\sin \frac{5x}{4}=1\Rightarrow \frac{5x}{4}=2n\pi +\frac{\pi }{2}\Rightarrow x=\frac{8n\pi }{5}+\frac{2\pi }{5}\] & \[\cos x=1\Rightarrow x=2m\pi \] Thus we have \[\frac{8n\pi }{5}+\frac{2\pi }{5}=2m\pi \,\,\,\Rightarrow \,\,m=\frac{4n+1}{5}\] \[\therefore \,\,\,n\in I,\] so m must be of the form \[m=5k+1\] Hence the solution of the equation is \[x=2\,(5k+1)\,\pi ,\]\[k\in I\]
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