solve the trig equation

We note that the right-hand side, $2^x + 2^{-x} > 0$ , therefore the left-hand side, $2\cos \left(\dfrac {x^2+x}6\right)$ must also be $>0$ . Note that the right-hand side has a minimum value of $2$ , when $x=0$ , while the left-hand side has a maximum value of $2$ , also when $x=0$ . Since $2^x + 2^{-x} = 2$ is unique, the solution $x=\boxed 0$ is also unique.

Note that the minimum value of $2^x + 2^{-x}$ is $2$ (from AM-GM), which occurs at $x = 0$ .

Note that the maximum value of $2 \cos\left(\displaystyle \frac{x^2+x}{6}\right)$ is $2$ . Because $x = 0$ is a value that makes $2 \cos\left(\displaystyle \frac{x^2+x}{6}\right)$ equal to 2, our answer is $x = \boxed{0}$ .