https://file.aiquickdraw.com/mathbot/file/1727328243826gxdcftlm.jpg

The problem presented in the image is:

Solve for \(x\) in the equation:

\[
\sin(x + 30^\circ) + 1 = 0
\]

with the condition \(0^\circ \leq x \leq 360^\circ\).

### Step-by-Step Solution:

1. **Isolate the sine function:**

\[
\sin(x + 30^\circ) + 1 = 0
\]
\[
\sin(x + 30^\circ) = -1
\]

2. **Determine where \(\sin(\theta) = -1\):**

The sine function equals \(-1\) at \( \theta = 270^\circ \) within one cycle \( [0^\circ, 360^\circ] \).

3. **Solve for \(x\):**

We have:

\[
x + 30^\circ = 270^\circ
\]

Now, solve for \(x\):

\[
x = 270^\circ - 30^\circ
\]
\[
x = 240^\circ
\]

4. **Check the solution:**

The solution satisfies the condition \(0^\circ \leq x \leq 360^\circ\).

Thus, the solution is:

\[
x = 240^\circ
\]

### Answer:
The correct answer is:

\[
\boxed{c. \{240^\circ\}}
\]

Would you like further clarification or additional details on this solution?

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Here are five related questions:
1. How can you solve trigonometric equations involving different angles like \(\sin(x + 45^\circ)\)?
2. How does the unit circle help in solving trigonometric equations?
3. What is the general solution for \(\sin(\theta) = -1\)?
4. How can this solution method be applied to other trigonometric functions like cosine or tangent?
5. Why is it important to check the range \(0^\circ \leq x \leq 360^\circ\) in this type of problem?

**Tip:** When solving trigonometric equations, always consider both the reference angle and the quadrant to ensure you capture all possible solutions within the given range.

Himpunan penyelesaian sin(x + 30°) + 1 = 0 dengan 0° ≤ x ≤ 360°

Learn how to solve the trigonometric equation sin(x + 30°) + 1 = 0 within the interval 0° ≤ x ≤ 360°. This problem involves basic trigonometric concepts and the sine function, suitable for students in Grades 10-12.

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