The problem asks you to find all solutions to the equation:

$\cos x = \sqrt{1 - \sin x}$

over the interval $[0, 2\pi]$.

Step-by-Step Solution

1. Square Both Sides: To eliminate the square root, square both sides: $\cos^2 x = 1 - \sin x$

2. Use the Pythagorean Identity: Recall that $\cos^2 x = 1 - \sin^2 x$. Substitute this identity: $1 - \sin^2 x = 1 - \sin x$

3. Simplify the Equation: Rearrange the equation by moving all terms involving $\sin x$ to one side: $\sin^2 x - \sin x = 0$

4. Factor the Equation: Factor out $\sin x$: $\sin x (\sin x - 1) = 0$

5. Solve for $\sin x$: This gives two cases:

   • Case 1: $\sin x = 0$
   • Case 2: $\sin x = 1$

6. Find $x$ for Each Case:

   • For $\sin x = 0$, $x = 0$ and $x = \pi$.
   • For $\sin x = 1$, $x = \frac{\pi}{2}$.

Final Answer

The solutions to the equation in the interval $[0, 2\pi]$ are: $x = 0, \frac{\pi}{2}, \pi$

These values should be entered as $0, \frac{\pi}{2}, \pi$ in terms of $\pi$.

Would you like a more detailed breakdown of any of these steps?

Here are 5 related questions for further exploration:

1. How would the solutions change if the interval were different, such as $[0, \pi]$?
2. What if the equation were $\cos x = -\sqrt{1 - \sin x}$ instead? How would that impact the solution?
3. Why is squaring both sides of an equation a valid approach, and what precautions should you take when using it?
4. What are the general solutions for $\sin x = 0$ and $\sin x = 1$ beyond the interval $[0, 2\pi]$?
5. How does the Pythagorean identity relate to other trigonometric identities?

Tip: When solving trigonometric equations, always check for extraneous solutions, especially when squaring both sides.