Trigonometric Equations Question 371
Question: If $ \cos 2\theta =(\sqrt{2}+1)( \cos \theta -\frac{1}{\sqrt{2}} ) $ , then the value of $ \theta $ is
[Roorkee 1977]
Options:
A) $ 2n\pi +\frac{\pi }{4} $
B) $ 2n\pi \pm \frac{\pi }{4} $
C) $ 2n\pi -\frac{\pi }{4} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ 2{{\cos }^{2}}\theta -(\sqrt{2}+1)\cos \theta -1+\frac{(\sqrt{2}+1)}{\sqrt{2}}=0 $
$ \Rightarrow $ $ \cos \theta =\frac{(\sqrt{2}+1)\pm \sqrt{{{(\sqrt{2}+1)}^{2}}-\frac{8}{\sqrt{2}}}}{4} $
$ \Rightarrow $ $ \cos \theta =\cos ( \frac{\pi }{4} ) $
$ \Rightarrow $ $ \theta =2n\pi \pm \frac{\pi }{4} $ . Trick: Since $ \theta =\frac{\pi }{4} $ satisfies the equation and therefore the general value should be $ 2n\pi \pm \frac{\pi }{4} $ .
BETA
