Trigonometric Identities Question 284
Question: Number of values of x which lie in $ [0,2\pi ] $ and satisfy the equation $ ( \cos \frac{x}{4}-2\sin x )\sin x+( 1+\sin \frac{x}{4}-2\cos x )\cos x=0 $
Options:
1
2
3
4
Show Answer
Answer:
Correct Answer: A
Solution:
$ ( \cos \frac{x}{2}-2\sin x )\sin x+( 1+\sin \frac{x}{4}-2\cos x )\cos x=0 $
$ \Rightarrow ( \sin x\cos \frac{x}{4}+\cos x\sin \frac{x}{4} )+\cos x-2( {{\sin }^{2}}x+{{\cos }^{2}}x )=0 $
$ \Rightarrow \sin ( x+\frac{x}{4} )+\cos x-2=0\Rightarrow \sin \frac{5x}{4}+\cos x=2 $
$ \Rightarrow \sin \frac{5x}{4}=\cos x=0 $
$ \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 \mathbb{Z}, $ so m must be of the form $ m=5k+1 $ Hence the solution of the equation is $ x=2,(5k+1),\pi , $ $ k\in \mathbb{Z} $
BETA
