Trigonometric Equations Question 54
Question: $ 2{{\sin }^{2}}x+{{\sin }^{2}}2x=2,,-\pi <x<\pi , $ then $ x= $
[ISM Dhanbad 1989]
Options:
A) $ \pm \frac{\pi }{6} $
B) $ \pm \frac{\pi }{4} $
C) $ \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 ,\text{ or}(2n+1)\pi $
$ \Rightarrow $ $ x=(2n+1)\frac{\pi }{4}\text{ or }(2n+1)\frac{\pi }{2} $ Now put $ n=-2,,-1,,0,,1,,2 $
$ \therefore $ $ x=\frac{-3\pi }{4},,\frac{-\pi }{4},,\frac{\pi }{4},,\frac{3\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{\pi }{2},,\pm \frac{3\pi }{4} $ only.
BETA
