The equation $$2\,{\cos ^2}\frac{x}{2}{\sin ^2}x = {x^2} + {x^{ - 2}};0 < x \leqslant \frac{\pi }{2}$$        has

Solution :
The given equation is
$$\eqalign{ & 2\,{\cos ^2}\left( {\frac{x}{2}} \right){\sin ^2}x = {x^2} + \frac{1}{{{x^2}}}\,\,{\text{where }}0 < x \leqslant \frac{\pi }{2} \cr & {\text{L.H.S.}} = 2\,{\cos ^2}\frac{x}{2}{\sin ^2}x = \left( {1 + \cos x} \right){\sin ^2}x \cr & \because \,\,1 + \cos x < 2\,\,{\text{and }}{\sin ^2}x \leqslant 1\,\,{\text{for }}0 < x \leqslant \frac{\pi }{2} \cr & \therefore \,\,\left( {1 + \cos x} \right){\sin ^2}x < 2 \cr & {\text{And R}}{\text{.H}}{\text{.S}}{\text{. }} = {x^2} + \frac{1}{{{x^2}}} \geqslant 2 \cr & \therefore \,\,{\text{For }}0 < x \leqslant \frac{\pi }{2}. \cr} $$
given equation is not possible for any real value of $$x.$$