To solve the equation \( \sin(x) + \cos(x) = 1 \), we can use the following steps:

1. Square both sides of the equation:
   \[
   (\sin(x) + \cos(x))^2 = 1^2
   \]
   
2. Expand the left-hand side:
   \[
   \sin^2(x) + 2\sin(x)\cos(x) + \cos^2(x) = 1
   \]
   
3. Use the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \):
   \[
   1 + 2\sin(x)\cos(x) = 1
   \]
   
4. Subtract 1 from both sides:
   \[
   2\sin(x)\cos(x) = 0
   \]
   
5. Since \( 2\sin(x)\cos(x) = \sin(2x) \), we have:
   \[
   \sin(2x) = 0
   \]

6. Solve for \( 2x \):
   \[
   2x = n\pi \quad \text{where} \quad n \in \mathbb{Z}
   \]

7. Solve for \( x \):
   \[
   x = \frac{n\pi}{2}
   \]

Therefore, the solutions to the equation \( \sin(x) + \cos(x) = 1 \) are:
\[ 
x = \frac{n\pi}{2} \quad \text{where} \quad n \in \mathbb{Z}
\]

Would you like any more details or have any questions?

Here are some related questions you might consider:
1. How do you solve \( \sin(x) - \cos(x) = 0 \)?
2. What are the solutions for \( \sin^2(x) + \cos^2(x) = 1 \)?
3. How do you solve \( \tan(x) = 1 \)?
4. What is the general solution to \( \sin(x) = 0.5 \)?
5. How do you solve \( \cos(2x) = \sin(x) \)?

**Tip:** Always check for extraneous solutions when squaring both sides of an equation, as this can sometimes introduce solutions that do not satisfy the original equation.

Learn how to solve the trigonometric equation sin(x) + cos(x) = 1 step-by-step. This problem involves using the Pythagorean identity and the double angle formula. Suitable for students in Grades 10-12.

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