Question

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

• A. no real solution
• B. one real solution
• C. more than one solution
• D. none of these

Answer 1

The correct option is A

no real solution

Explanation for the correct option:

The given equation is as follows:

$2{\cos }^2\frac{x}{2}{\sin }^{2}x=x^2+x^{-2};0<x\le \frac{\mathrm{\pi }}{2}$

Consider the LHS=$\left(1+\cos x\right)\left({\sin }^{2}x\right)$ using $\cos 2x=2{\cos }^{2}x-1$

$$\begin{gathered} 0<x\le \frac{\mathrm{\pi }}{2} \\ 0\le \cos x<1 \\ 1\le 1+\cos x<2 \end{gathered}$$

And

$0<{\sin }^{2}x\le 1$

This gives:

$0\le \left(1+\cos x\right)\left({\sin }^{2}x\right)<2$

Now consider the RHS:

$$\begin{gathered} x^2+\frac{1}{x^2}={\left(x-\frac{1}{x}\right)}^2+2 \\ 0<{\left(x-\frac{1}{x}\right)}^2<\infty \\ 2<{\left(x-\frac{1}{x}\right)}^2+2<\infty \end{gathered}$$

Since LHS is between $[0,2)$ and RHS is between $(2,\infty )$there cannot be a solution satisfying the equation

Hence, the correct answer is option (A).