A) 0
B) 1
C) infinite
D) None of these
Correct Answer: A
Solution :
The given equation can be written as \[{{\sin }^{5}}x-{{\cos }^{5}}x=\frac{\sin x-\cos x}{\sin x\cos x}\] \[\Rightarrow \,\,\,\sin x\cos x\left[ \frac{{{\sin }^{5}}x-{{\cos }^{5}}x}{\sin x-\cos x} \right]=1\] \[\Rightarrow \,\,\frac{1}{2}\sin 2x[{{\sin }^{4}}x+{{\sin }^{3}}x\cos x+{{\sin }^{2}}x{{\cos }^{2}}x\] \[+\sin x{{\cos }^{3}}x+{{\cos }^{4}}x]=1\] \[\Rightarrow \,\,\,\sin 2x[{{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x\] \[+\sin x\cos x({{\sin }^{2}}x+{{\cos }^{2}}x)+{{\sin }^{2}}x{{\cos }^{2}}x]=2\] \[\Rightarrow \,\,\sin 2x[1-{{\sin }^{2}}x{{\cos }^{2}}x+\sin x\cos x]=2\] \[\Rightarrow \,\,\sin 2x\left[ 1-\frac{1}{4}{{\sin }^{2}}2x+\frac{1}{2}\sin 2x \right]=2\] \[\Rightarrow \,\,\,{{\sin }^{3}}2x-2{{\sin }^{2}}2x-4\sin 2x+8=0\] \[\Rightarrow \,\,{{(\sin 2x-2)}^{2}}(\sin 2x+2)=0\] \[\Rightarrow \,\,\sin 2x=\pm 2,\] which is not possible for any x.
You need to login to perform this action.
You will be redirected in
3 sec
