Answer: $A$
Minimum value of $2^x + 2^{-x} = 2$
$$\implies 2\cos\bigg(\frac{x^2+x}{6}\bigg) = 2^x + 2^{-x}$$
$$\implies 2\cos\bigg(\frac{x^2+x}{6}\bigg) \ge 2$$
Now, $\because \cos 0 = 1$
$$\implies \frac{x^2+x}{6} = 0 \implies x(x+1) = 0 \implies x = 0, x = -1$$
$$
So, $x = 0$ is the only solution as the equation is not satisfying at $x = -1$