A) one real solution
B) no solution
C) more than one real solution
D) None of these
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.
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