Number of real solutions $2\cos(\frac{x^2+x}{2})=2^x+2^{-x} $ The number of real solutions of $2\cos(\frac{x^2+x}{2})=2^x+2^{-x} $  is 
(1) 0
(2) 1 
(3) 2
(4) infinitely many .
My work : 
$$
1\geq \cos\left(\frac{x^2+x}{2}\right)=\frac{2^x+2^{-x} }{2}\geq 1 
\qquad \text{by (AM-GM).}
$$
So $\frac{x^2+x}{2}=2n\pi$ for all $n\in \mathbb{Z}$ . Now discriminant $=1+2n\pi$  is always positive for $n\geq0$  . But the equation is a quadratic so it has only two solution . Hence the answer must be 2 .
PS: I'm aware that the this problem is already on the site but i posted as there were no complete solution .

hint
$$-2\leq 2\cos(\frac {x^2+x}{2})\leq 2$$
$$2^x+2^{-x}\geq 2$$
the root must satisfy
$$2^x+2^{-x}=2$$ which gives $x=0$.
the unique root.
