Trigonometric Equations Question 72
Question: Common roots of the equations $ 2{{\sin }^{2}}x+{{\sin }^{2}}2x=2 $ and $ \sin 2x+\cos 2x=\tan x, $ are
Options:
A) $ x=(2n-1)\frac{\pi }{2} $
B) $ x=(2n+1)\frac{\pi }{4} $
C) $ x=(2n+1)\frac{\pi }{3} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ 2{{\sin }^{2}}x+{{\sin }^{2}}2x=2 $ ……(i) and $ \sin 2x+\cos 2x=\tan x $ …..(ii) Solving (i), $ {{\sin }^{2}}2x=2{{\cos }^{2}}x $
Þ $ 2{{\cos }^{2}}x\cos 2x=0 $ Þ $ x=(2n+1)\frac{\pi }{2}\text{ or }x=(2n+1)\frac{\pi }{4} $
$ \therefore $ Common roots are $ (2n\pm 1)\frac{\pi }{4} $ Solving (ii), $ \frac{2\tan x+1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\tan x $
$ \Rightarrow $ $ {{\tan }^{3}}x+{{\tan }^{2}}x-\tan x-1=0 $
$ \Rightarrow $ $ ({{\tan }^{2}}x-1),(\tan x+1)=0 $
$ \Rightarrow $ $ x=m\pi \pm \frac{\pi }{4} $ Trick: For $ n=0 $ , option gives $ \theta =-\frac{\pi }{2} $ which satisfies the equation (i) but does not satisfy the (ii). Now option gives $ \theta =\frac{\pi }{4} $ which satisfies both the equations.
BETA
