Trigonometric Equations Question 14
Question: If $ 4{{\sin }^{4}}x+{{\cos }^{4}}x=1, $ then x =
[Roorkee 1989]
Options:
A) $ n\pi $
B) $ n\pi \pm {{\sin }^{-1}}\frac{2}{5} $
C) $ n\pi +\frac{\pi }{6} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- The given equation can be put in the form $ 4{{\sin }^{4}}x=1-{{\cos }^{4}}x=(1-{{\cos }^{2}}x),(1+{{\cos }^{2}}x) $
$ \Rightarrow $ $ {{\sin }^{2}}x[4{{\sin }^{2}}x-1-(1-{{\sin }^{2}}x)]=0 $
$ \Rightarrow $ $ {{\sin }^{2}}x[5{{\sin }^{2}}x-2]=0 $
$ \Rightarrow $ $ \sin x=0 $ or $ \sin x=\pm \sqrt{2/5} $ . Hence $ x=n\pi $ is the required answer.
BETA
