My favourite technique: LHS and RHS are both even functions so that the answer is $\boxed{0}$ by symmetry

By the AM-GM inequality we have that $2^{x} + \dfrac{1}{2^{x}} \ge 2\sqrt{2^{x}*\dfrac{1}{2^{x}}} = 2$ ,

with equality holding when $2^{x} = \dfrac{1}{2^{x}} \Longrightarrow 2^{2x} = 1 \Longrightarrow x = 0$ .

Now for $x \in \mathbb{R}$ we have $-2 \le 2\cos(x) \le 2$ , so the given equality holds only when both sides equal $2$ , which only occurs when $x = \boxed{0}$ .

$2^x+2^{-x}=e^{\ln\left(2\right)x}+e^{-\ln\left(2\right)x}=2\cosh(\ln(2)\cdot x)$

$\cosh(\ln(2)\cdot x)=\cos(i\ln(2)\cdot x)$

Now, we have $\cos(x)=\cos(i\ln(2)\cdot x)$ and $x=i\ln(2)\cdot x+k\pi$

Because x is real, the only solution is $\boxed{0}$

We can use the fact that, max & min values of cosx is -1 & 1 (for real x) and also in our case, cosx can't take negative values which makes the problem more simple!