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.

Notice that $2^0+2^0=2$ and that $2^x+2^{-x}> 2$ for $x\ne 0$.  Then consider that $2\cos(0)=2$ and that $2\cos\left(\frac{x^2+x}{6}\right)\le2$ for all $x$.  What can we conclude?
