Trigonometric Equations Question 15
Question: If $ \cos 3x+\sin ( 2x-\frac{7\pi }{6} )=-2 $ , then $ x= $ (where $ k\in Z $ )
Options:
A) $ \frac{\pi }{3}(6k+1) $
B) $ \frac{\pi }{3}(6k+1) $
C) $ \frac{\pi }{3}(2k+1) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- We have $ \cos 3x+\sin ( 2x-\frac{7\pi }{6} ),=-2 $
$ \Rightarrow $ $ 1+\cos 3x+1+\sin ( 2x-\frac{7\pi }{6} )=0 $
$ \Rightarrow $ $ (1+\cos 3x)+1-\cos ( 2x-\frac{2\pi }{3} )=0 $
$ \Rightarrow $ $ 2{{\cos }^{2}}\frac{3x}{2}+2{{\sin }^{2}}( x-\frac{\pi }{3} )=0 $
$ \Rightarrow $ $ \cos \frac{3x}{2}=0 $ and $ \sin ( x-\frac{\pi }{3} )=0 $
$ \Rightarrow $ $ \frac{3x}{2}=\frac{\pi }{2},,\frac{3\pi }{2},,….. $ and $ x-\frac{\pi }{3} $ =0, $ \pi ,2\pi …..\Rightarrow x=\frac{\pi }{3} $ Therefore, the general solution of $ \cos \frac{3x}{2}=0 $ and $ \sin ( x-\frac{\pi }{3} )=0 $ is $ x=2k\pi +\frac{\pi }{3}=\frac{\pi }{3}(6k+1), $ where $ k\in Z $ .
BETA
