Number of real roots of $2 \cos\left(\frac{x^2+x}{6}\right)=2^x+2^{-x}$ Find the number of real roots of 
$ \cos \,\left(\dfrac{x^2+x}{6}\right)= \dfrac{2^x+2^{-x}}{2}$
1) 0
2) 1
3) 2
4)  None of these
My guess is to approach it in graphical way. But equation seems little difficult.

Rearranging we get $$(2^x)^2-2\cos\dfrac{x^2+x}6(2^x)+1=0\  \ \ \ (0)$$ which is a Quadratic equation in $2^x$
For real $2^x,$ the discriminant must be $\ge0$
i.e., $$\left(2\cos\dfrac{x^2+x}6\right)^2-4=-4\sin^2\dfrac{x^2+x}6\ge0\  \ \ \ (1)$$
As $\sin\dfrac{x^2+x}6$ is real, $$\sin^2\dfrac{x^2+x}6\ge0\iff-4\sin^2\dfrac{x^2+x}6\le0\ \ \ \ (2)$$
By $(1),(2);$  $$\sin^2\dfrac{x^2+x}6=0\implies\cos\dfrac{x^2+x}6=\pm1$$
What happens to $(0)$ if $(i):\cos\dfrac{x^2+x}6=1$
Or if $(ii):\cos\dfrac{x^2+x}6=-1$
Remember for real $x, 2^x>0$
