JEE Main 2016 (6)

Checking the root for each factor:

Case 1:

$(2^{x}+2^{-x}-2\cos(x)) = 0$

$\implies 2^{x}+2^{-x} = 2\cos(x))$

Using AM $\ge$ GM: $2^{x}+2^{-x} \ge 2$

But $\cos(x) \in [-1, 1]$

$\therefore 2\cos(x) \in [-2, 2]$

This shows that the ranges of LHS and RHS only have ${2}$ in common.

As the equality in AM-GM is achieved when both the terms are equal, $x=0$ and it can be seen that at $x = 0$ , 2 $\cos(x)$ is indeed 2.

$\therefore x=0$ is one root.

Case 2:

$(3^{\pi+x}+3^{-x-\pi}+2\cos(x)) = 0$

By similar argument as above, $x=-\pi$ must be the solution and it does satisfy the equation.

$\therefore x=-\pi$ is one root.

Case 3:

$(5^{\pi-x}+5^{-\pi+x}-2\cos(x))=0$

By similar argument as above, $x = \pi$ must be a solution but it does not satisfy the equation.

$\therefore x=0, -\pi$ are the only roots.