<s>[INST] 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.
 [/INST] If you consider $f(x)=2^x+2^{-x}$ and notice how it is symmetrical around $x=0$ and has its minimum there as for example $f(1)=2.5$ and $f(2)=4.25$, which the left hand side has a maximum value of 2, thus there is only one possible root though I'll leave that for you to see.
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