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.

There's a clever approach to this problem. In particular, $$(w-1)^2 \geq 0 \implies w^2 + 1 \geq 2w \implies w + \frac{1}{w} \geq 2$$
and this works for any $w$. Using this, you can show that one side is always at least $2$. Meanwhile, the other is at most $2$.
