How to find the number of real roots of the given equation? 
The number of real roots of the equation $$2 \cos \left( \frac{x^2+x}{6} \right)=2^x+2^{-x}$$ is
  (A) $0$, (B) $1$, (C) $2$, (D) infinitely many.

Trial:  $$\begin{align} 2 \cos \left( \frac{x^2+x}{6} \right)&=2^x+2^{-x} \\ \implies \frac{x^2+x}{6}&=\cos ^{-1} \left( \frac{2^x+2^{-x}}{2} \right)\end{align}$$ Then I can't proceed.

*

*$RHS=2^x+2^{-x}=2^x+\frac{1}{2^x}\geq 2$ by $AM\geq GM$, since $2^x>0$ for all $x\in\mathbb{R}$

*$LHS\leq 2$ for all $x\in\mathbb{R}$. This is obvious.

*One can verify $x=0$  is a solution. 

*$f(x)=2^x+2^{-x}$ implies $f'(x)>0$ for $x>0$. 
These should suffice to conclude the unique solution at $x=0$. 
