The equation 2 cos ^{2} left( frac{x}{2} righ | vedclass

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(A) The given equation is $2 \cos ^{2} \left( \frac{x}{2} \right) \sin ^{2} x = x^{2} + \frac{1}{x^{2}}$.For the $R.H.S.$,by $AM \geq GM$ inequality,$

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no solution

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(A) The given equation is $2 \cos ^{2} \left( \frac{x}{2} ight) \sin ^{2} x = x^{2} + \frac{1}{x^{2}}$.For the $R.H.S.$,by $AM \geq GM$ inequality,$x^{2} + \frac{1}{x^{2}} \geq 2$ for all $x 
eq 0$. The equality holds when $x^{2} = 1$,i.e.,$x = \pm 1$.For the $L.H.S.$,we have $2 \cos ^{2} \left( \frac{x}{2} ight) \sin ^{2} x$. Since $\cos ^{2} \left( \frac{x}{2} ight) \leq 1$ and $\sin ^{2} x \leq 1$,the maximum value of the product $2 \cos ^{2} \left( \frac{x}{2} ight) \sin ^{2} x$ is $2 	imes 1 	imes 1 = 2$.For the equation to hold,both sides must equal $2$. This requires $x^{2} = 1$ (so $x = 1$ since $x \in [0, \pi/2]$) $AND$ $\cos ^{2} \left( \frac{x}{2} ight) = 1$ and $\sin ^{2} x = 1$.If $x = 1$,then $\sin ^{2} (1) \approx (0.841)^{2} \approx 0.707 
eq 1$.Since the $L.H.S.$ cannot reach $2$ when the $R.H.S.$ is $2$,there is no solution.

The equation $2 \cos ^{2} \left( \frac{x}{2} \right) \sin ^{2} x = x^{2} + \frac{1}{x^{2}}$ for $0 \leq x \leq \frac{\pi}{2}$ has

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