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The problem is to solve the equation:

\[
\sin(x) = \cos(x) + 1
\]

for \( x \) within the interval \([0, 2\pi)\) and express the answers in radians in terms of \(\pi\).

### Step-by-Step Solution:

1. **Rewrite the Equation**: Start by rearranging terms to isolate \(\sin(x)\) and \(\cos(x)\):
   \[
   \sin(x) - \cos(x) = 1
   \]

2. **Square Both Sides**: To simplify trigonometric equations involving sine and cosine, a common technique is to square both sides. However, since this may introduce extraneous solutions, we’ll need to check solutions at the end:
   \[
   (\sin(x) - \cos(x))^2 = 1^2
   \]
   Expanding both sides:
   \[
   \sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) = 1
   \]

3. **Use a Pythagorean Identity**: Recall that \(\sin^2(x) + \cos^2(x) = 1\), so we can substitute \(1\) for \(\sin^2(x) + \cos^2(x)\) in the equation:
   \[
   1 - 2\sin(x)\cos(x) = 1
   \]
   Simplify by subtracting \(1\) from both sides:
   \[
   -2\sin(x)\cos(x) = 0
   \]

4. **Factor the Equation**: We can now factor out \(-2\):
   \[
   \sin(x)\cos(x) = 0
   \]
   
   This implies that either \(\sin(x) = 0\) or \(\cos(x) = 0\).

5. **Solve for \(x\)**:
   - If \(\sin(x) = 0\), then \(x = 0\) or \(x = \pi\) (within the interval \([0, 2\pi)\)).
   - If \(\cos(x) = 0\), then \(x = \frac{\pi}{2}\) or \(x = \frac{3\pi}{2}\) (within the interval \([0, 2\pi)\)).

6. **Verify Solutions**: Substitute each value back into the original equation to ensure it satisfies \(\sin(x) = \cos(x) + 1\).
   - For \(x = 0\): \(\sin(0) = 0\) and \(\cos(0) + 1 = 1\), which does not satisfy the equation.
   - For \(x = \pi\): \(\sin(\pi) = 0\) and \(\cos(\pi) + 1 = 0\), which does satisfy the equation.
   - For \(x = \frac{\pi}{2}\): \(\sin\left(\frac{\pi}{2}\right) = 1\) and \(\cos\left(\frac{\pi}{2}\right) + 1 = 1\), which does satisfy the equation.
   - For \(x = \frac{3\pi}{2}\): \(\sin\left(\frac{3\pi}{2}\right) = -1\) and \(\cos\left(\frac{3\pi}{2}\right) + 1 = 0\), which does not satisfy the equation.

### Solution:
The values of \(x\) that satisfy the equation in the interval \([0, 2\pi)\) are:
\[
x = \pi, \frac{\pi}{2}
\]

### Final Answer:
\[
x = \frac{\pi}{2}, \pi
\]

Would you like more details on this solution, or do you have any questions?

Here are some related questions:
1. How do we handle extraneous solutions when squaring equations?
2. What are alternative methods for solving trigonometric equations involving sine and cosine?
3. Can we solve similar problems without using identities?
4. Why is verifying solutions critical in trigonometric equations?
5. How does the unit circle help in understanding trigonometric solutions?

**Tip:** Always verify each solution after squaring both sides of an equation, as this can introduce extraneous solutions.

Find all solutions of the equation sin(x) = cos(x) + 1 in the interval [0, 2π).

Learn how to solve the trigonometric equation sin(x) = cos(x) + 1 within the interval [0, 2π) using Pythagorean identities. This guide provides a step-by-step solution, including factoring techniques and verification of solutions. Suitable for high school students in Grades 10-12.

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