Answer
$$\frac{3 \sqrt{3}}{4}$$
Work Step by Step
Given
$$y=\cos x, \quad y=\cos 2 x, \quad x=0, \quad x=\frac{2 \pi}{3}$$
Since
\begin{align*}
\cos x&=\cos2x\\
\cos x&=\cos^2x-\sin^2\\
2\cos^2x-\cos x-1&=0\\
(2 \cos x+1)(\cos x-1)&=0
\end{align*}
Then $x=0, \quad x=\frac{2 \pi}{3}$, and $ \cos x\geq\cos 2x$ for $ 0\leq x\leq 2\pi/3$
Hence,
\begin{aligned}
A &=\int_{0}^{\frac{2 \pi}{3}}(f(x)-g(x)) d x \\
&=\int_{0}^{\frac{2 \pi}{3}}(\cos x-\cos 2 x) d x \\
&=\left[\sin x-\frac{\sin 2 x}{2}\right]_{0}^{\frac{2 \pi}{3}} \\
&=\frac{3 \sqrt{3}}{4}
\end{aligned}
