Answer
$(0,1)$, $(\pi,1)$, $(2\pi,1)$, $(\frac{\pi}{6},\frac{1}{2})$ and $(\frac{5\pi}{6},\frac{1}{2})$; see graph.
Work Step by Step
Step 1. Graph $f(x)$ (red curve) and $g(x)$ (blue curve) as shown in the figure.
Step 2. Let $f(x)=g(x)$; we have $cos(2x)=1-sin(x)$. Using $cos(2x)=1-2sin^2x$ we have $2sin^2x-sin(x)=0$, which gives $sin(x)=0$ and $sin(x)=\frac{1}{2}$
Step 3. For $sin(x)=0$, the solutions in $[0,2\pi]$ are $x=0,\pi,2\pi$ corresponding to points of intersection $(0,1)$, $(\pi,1)$, $(2\pi,1)$
Step 4. For $sin(x)=\frac{1}{2}$, we can find the solutions in $[0,2\pi]$ as $x=\frac{\pi}{6}, \frac{5\pi}{6} $ corresponding to points of intersection $(\frac{\pi}{6},\frac{1}{2})$ and $(\frac{5\pi}{6},\frac{1}{2})$
Step 5. We can identify the above points as shown on the graph.
