Let $x$ be an acute angle such that $\tan x + \sec x = 2 \cot x + \csc x.$ Find $\cos x$.
Dividing $\tan x + \sec x = 2 \cot x + \csc x$ by $\sin x$,we get
\[\frac{\sin x}{\cos x} + \frac{1}{\cos x} = \frac{2 \cos x}{\sin x} + \frac{1}{\sin x},\]or $1 + \cos x = 2 \sin x + \cos x.$ Hence, $\sin x = 1 - 2 \sin x$, so $3 \sin x = 1$. Then $\sin x = \frac{1}{3}$ and
\[\cos^2 x = 1 - \sin^2 x = \frac{8}{9},\]so $\cos x = \boxed{\frac{2 \sqrt{2}}{3}}.$