Trigonometric Identities Question 285
Question: The number of solutions of the equation $ {{\sin }^{5}}x-{{\cos }^{5}}x=\frac{1}{\cos x}-\frac{1}{sinx}(\sin x\ne \cos x) $ is
Options:
A) 0
B) 1
C) infinite
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
The given equation can be written as $ {{\sin }^{5}}x-{{\cos }^{5}}x=\frac{\sin x-\cos x}{\sin x\cos x} $
$ \Rightarrow ,\sin x\cos x[ \frac{{{\sin }^{5}}x-{{\cos }^{5}}x}{\sin x-\cos x} ]=1 $
$ \Rightarrow \frac{1}{2}\sin 2x[{{\sin }^{4}}x+{{\sin }^{3}}x\cos x+{{\sin }^{2}}x{{\cos }^{2}}x $ $ +\sin x{{\cos }^{3}}x+{{\cos }^{4}}x]=1 $
$ \Rightarrow ,\sin 2x[{{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x $ $ +\sin x\cos x({{\sin }^{2}}x+{{\cos }^{2}}x)+{{\sin }^{2}}x{{\cos }^{2}}x]=2 $
$ \Rightarrow \sin 2x[1-{{\sin }^{2}}x{{\cos }^{2}}x+\sin x\cos x]=2 $
$ \Rightarrow \sin 2x[ 1-\frac{1}{4}{{\sin }^{2}}2x+\frac{1}{2}\sin 2x ]=2 $
$ \Rightarrow ,{{\sin }^{3}}2x-2{{\sin }^{2}}2x-4\sin 2x+8=0 $
$ \Rightarrow {{(\sin 2x-2)}^{2}}(\sin 2x+2)=0 $
$ \Rightarrow \sin 2x=\pm 2, $ which is not possible for any x.
BETA
