Step 1: Understand what is asked.
We must count how many angles $\theta$ between $0$ and $2\pi$ make the equation $\cos 2\theta = \sin\theta$ true. The interval is open, so $0$ and $2\pi$ are not counted.
Step 2: Make both sides use the same trig function.
Mixing $\cos 2\theta$ and $\sin\theta$ is hard. So we rewrite $\cos 2\theta$ using only $\sin\theta$. The standard double angle form is $\cos 2\theta = 1 - 2\sin^2\theta$.
Step 3: Form a simple equation.
Putting this in gives \[ 1 - 2\sin^2\theta = \sin\theta. \] Bring all terms to one side so it looks like a normal quadratic. \[ 2\sin^2\theta + \sin\theta - 1 = 0. \]
Step 4: Treat it as a quadratic in $\sin\theta$.
Let $s = \sin\theta$. Then $2s^2 + s - 1 = 0$. This factors neatly as \[ (2s - 1)(s + 1) = 0. \] So either $s = \tfrac{1}{2}$ or $s = -1$.
Step 5: Find the angles for each value.
For $\sin\theta = \tfrac{1}{2}$, the angles in $(0,2\pi)$ are $\theta = \tfrac{\pi}{6}$ and $\theta = \tfrac{5\pi}{6}$ (sine is positive in the first and second quadrants). For $\sin\theta = -1$, the only angle is $\theta = \tfrac{3\pi}{2}$.
Step 6: Count the valid solutions.
All three angles $\tfrac{\pi}{6}$, $\tfrac{5\pi}{6}$ and $\tfrac{3\pi}{2}$ lie inside $(0,2\pi)$, so the count is $3$. \[ \boxed{3 \text{ solutions}} \]