Trigonometric Identities Question 274
Question: The equation $ 2{{\cos }^{2}}( \frac{x}{2} ).{{\sin }^{2}}x=x^{2}+\frac{1}{x^{2}}, $ $ 0\le x\le \frac{\pi }{2} $ has
Options:
A) one real solution
B) no solution
C) more than one real solution
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Since $ x^{2}+{x^{-2}}={{(x-{x^{-1}})}^{2}}+2\le 2 $ and $ 2{{\cos }^{2}}\frac{x}{2}{{\sin }^{2}}x\le 2, $
$ \therefore $ the given equation is valid only if $ 2{{\cos }^{2}}\frac{x}{2}{{\sin }^{2}}x=2 $ $ \Leftrightarrow \cos \frac{x}{2}=\cos ,ecx=1, $ which cannot be true.
BETA
