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Q.
The equation $2 \cos ^{2} \frac{x}{2} \sin ^{2} x=x^{2}+\frac{1}{x^{2}}, 0 \leq x \leq \frac{\pi}{2}$ has
Complex Numbers and Quadratic Equations
Solution:
We have for any real $x, \cos ^{2} \frac{x}{2} \leq 1$ and $\sin ^{2} x \leq 1$
$\therefore 2 \cos ^{2} \frac{x}{2} \sin ^{2} x \leq 2$
$ \Rightarrow $ LHS $\leq 2$
Again for any real $x$
$x ^{2}+\frac{1}{ x ^{2}}= x ^{2}+\frac{1}{ x ^{2}}-2+2=\left( x -\frac{1}{ x }\right)^{2}+2 \geq 2$
$\therefore RHS \geq 2$ Thus, the given equation can have solution only if $L H S=RHS=2$
$\Rightarrow 2 \cos ^{2} \frac{x}{2} \sin ^{2} x=x^{2}+\frac{1}{x^{2}}=2$
Now, $x^{2}+\frac{1}{x^{2}}=2$
$ \Rightarrow x=\pm 1$
but for $x=\pm 1, \cos ^{2} \frac{x}{2} \sin ^{2} x \neq 1$
That is, LHS and RHS cannot be simultaneously equal to 2 for any value of $x$.
$\therefore $ The equation has no real solution.