Find exactly all solutions to the given equation.
\[
2 \cos ^{2} \theta=-3 \sin \theta+3,0 \leq \theta \leq 2 \pi
\]

Step 1 ：Given the equation \(2 \cos ^{2} \theta=-3 \sin \theta+3\), where \(0 \leq \theta \leq 2 \pi\).

Step 2 ：We can simplify this equation by using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\).

Step 3 ：Rewrite this identity as \(\cos^2\theta = 1 - \sin^2\theta\) and substitute this into the given equation to get an equation in terms of sine only.

Step 4 ：Solving this equation for \(\sin\theta\), we get the solutions as \(\sin\theta = \frac{1}{2}\) and \(\sin\theta = 1\).

Step 5 ：We need to find the corresponding values of \(\theta\) in the range \([0, 2\pi]\). We can use the arcsin function to find these values.

Step 6 ：However, the arcsin function only gives values in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\). We need to consider the unit circle to find all possible values of \(\theta\).

Step 7 ：By considering the unit circle, we find the values of \(\theta\) that correspond to \(\sin\theta = \frac{1}{2}\) and \(\sin\theta = 1\) are \(\theta = 0.524\) and \(\theta = 1.571\) respectively.

Step 8 ：Final Answer: The solutions to the given equation in the range \([0, 2\pi]\) are \(\boxed{0.524}\) and \(\boxed{1.571}\).