The equation $$2\,{\cos ^2}\left( {\frac{x}{2}} \right) \cdot {\sin ^2}x = {x^2} + \frac{1}{{{x^2}}},0 \leqslant x \leqslant \frac{\pi }{2}{\text{ has}}$$

Solution :
$$\eqalign{ & {\text{Since, }}{x^2} + {x^{ - 2}} = {\left( {x - {x^{ - 1}}} \right)^2} + 2 \leqslant 2 \cr & {\text{and }}2\,{\cos ^2}\frac{x}{2}{\sin ^2}x \leqslant 2, \cr} $$
$$\therefore $$ the given equation is valid only if
$$2\,{\cos ^2}\frac{x}{2}{\sin ^2}x = 2$$
$$ \Leftrightarrow \cos \frac{x}{2} = \operatorname{cosec} x = 1,$$     which cannot be true.