find number of real roots

Question

The number of real roots of the equation $$2\cos((x^2+x)/6) = 2^x + 2^{−x}$$

My approach: If I put $x=0$ in both side then $LHS=RHS$ so one real solution is zero

but I'm not able to find if it has any other real root or not

Answer 1

By the MA-MG inequality, $\cos\left(\frac{x^2+x}{6}\right)=\frac{2^x+\frac{1}{2^x}}{2}\ge 1$, with equality iff $2^x=\frac{1}{2^x}$ iff $x=0$. Then, for $x\neq 0$, the LHS of the inequality is $>1$.

But $\cos\theta\le 1$. Thus $x=0$ is the only one solution.

Answer 2

Hint: Show that if $x\neq 0$, then the right-hand side is always greater than $2$.