Question

The number of real solutions $x$ of the equation
${\cos }^2\left(x\sin \right(2x\left)\right)+\frac{1}{1+x^2}={\cos }^{2}x+{\sec }^{2}x$

• A. 0.0
• B. 1
• C. 2
• D. infinite

Answer 1

The correct option is B 1

We know that
${\cos }^{2}x+{\sec }^{2}x\ge 2$........(i)
$0<\frac{1}{1+x^2}\le 1$ and $0\le {\cos }^2\left(x\sin 2x\right)\le 1$
so $0<\left({\cos }^2\right(x\sin \left(2x\right))+\frac{1}{1+x^2})\le 2$....(ii)
So equation (i) and (ii) will be equal when both are equal to 2.
$\therefore {\cos }^{2}x+{\sec }^{2}x=2\Rightarrow x=0$

Hence the only one possible solution will exist at $x=0$.