Rewriting $\sin 2x = \sin x \cos x + \cos x \sin x = 2\sin x\cos x$ we can compute the intersection: $\cos x = \sin(2x)$ is the same as $\begin{align*} \cos x&= 2\sin x\cos x\\ \cos x - 2\sin x\cos x &= 0\\ \cos x(1 - 2\sin x) &= 0. \end{align*}$ The product is zero if and only if $\cos x = 0$ (which on $[0,\pi/2]$ occurs only at $x=\pi/2$), or if $1-2\sin x = 0$, which is the same as $2\sin x = 1$, which is the same as $\sin x = \frac{1}{2}$; on $[0,\pi/2]$, this happens once and only once: at $x=\pi/6$.
So the point of intersection is at $x=\pi/6$.
On $[0,\pi/6]$, we have that $\cos(x)$ is greater than $\sin(2x)$. On $[\pi/6,\pi/2]$, we have that $\sin(2x)$ is greater than $\cos x$. So the area is given by $\begin{align*} \text{Area} &= \int_0^{\pi/2}|\cos x-\sin(2x)|\,dx\\ &= \int_0^{\pi/6}|\cos x - \sin(2x)|\,dx + \int_{\pi/6}^{\pi/2} |\cos x - \sin(2x)|\,dx\\ &= \int_0^{\pi/6}(\cos x - \sin (2x))\,dx + \int_{\pi/6}^{\pi/2}(\sin(2x) - \cos x)\,dx. \end{align*}$ Now you can simply compute the integrals and add up the appropriate quantities.