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.

$$\text{So, }(2^x)^2- 2^x\cdot2\cos\left(\frac{x^2+x}6\right)+1=0$$ which is a quadratic equation in $2^x$ 
As $x$ is real, so would be $2^x$ 
so,the discriminant must be $\ge 0$ 
i.e., $$\{2 \cos\left(\frac{x^2+x}6\right)\} ^2-4\cdot1\cdot1\ge0$$
$$\implies4\left(\cos^2\left(\frac{x^2+x}6\right)-1\right)\ge0$$
$$\implies  \sin^2\left(\frac{x^2+x}6\right)\le 0$$
$$\implies  \sin^2\left(\frac{x^2+x}6\right)= 0\text{ as the square of a real number }\ge 0 $$
So, the discriminant must be $0,\implies \cos\left(\frac{x^2+x}6\right)=\pm1$
$\implies 2^x=\frac{2\cos\left(\frac{x^2+x}6\right)\pm\sqrt{2 \cos\left(\frac{x^2+x}6\right)\}^2-4\cdot1\cdot1}}2=\pm1$
As $x$ is real, $2^x>0$
$\implies 2^x=1\implies x=0$ which also satisfies $ \sin^2\left(\frac{x^2+x}6\right)= 0$
