Finding the number of real solutions of $2 \cos((x^2+x)/6) = 2^x + 2^{-x}$ 
The number of real roots of the equation
  $$
 2\cos\left(\frac{x^2+x}{6}\right) = 2^x + 2^{-x}
$$
  is
  
  
*
  
*$0$,
  
*$1$,
  
*$2$,
  
*infinitely many.
  

I can see that $x=0$ satisfies the equation. But I do not know how to solve the equation explicitly.
Thanks

we have $$2\cos\left(\frac{x^2+x}{6}\right)=2^x+2^{-x}\geq 2\sqrt{2^x\cdot 2^{-x}}=2$$ by AM-GM thus we have $$2\cos\left(\frac{x^2+x}{6}\right)\geq 2$$
Can you proceed?
