Trigonometric Identities Question 266
Question: $ 2{{\sin }^{2}}x+{{\sin }^{2}}2x=2, $ $ -\pi <x<\pi , $ then; $ x= $
Options:
A) $ \pm \frac{\pi }{6} $
B) $ \pm \frac{\pi }{4} $
C) $ \pm \frac{3\pi }{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ 1-\cos 2x+1-{{\cos }^{2}}2x=2 $ or $ \cos 2x(\cos 2x+1)=0 $
$ \therefore \cos 2x=0,,-1, $
$ \therefore 2x=( n+\frac{1}{2} )\pi $ or $ (2n+1),\pi $
$ \Rightarrow ,x=(2n+1)\frac{\pi }{4} $ or $ (2n+1)\frac{\pi }{2} $ Now, put $ n=-2,-1,0,1,21 $
$ \therefore x=\frac{-3\pi }{4},\frac{-\pi }{4},\frac{\pi }{4},\frac{5\pi }{4} $ , and $ \frac{-3\pi }{2},\frac{-\pi }{2},\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2} $ Since $ -\pi \le x\le \pi , $ therefore, $ x=\pm \frac{\pi }{4},\pm \frac{3\pi }{4} $ only.
BETA
