https://r2.erweima.ai/mathbot/image/172707274472136lmvtor.jpg

The equation in the image is:

\[
\sin\left(x + \frac{\pi}{4}\right) = \frac{\sqrt{3}}{2}
\]

To solve for \(x\), let's proceed step-by-step:

1. The sine function equals \(\frac{\sqrt{3}}{2}\) at specific angles:
   \[
   \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}
   \]
   Therefore, we need:
   \[
   x + \frac{\pi}{4} = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad x + \frac{\pi}{4} = \pi - \frac{\pi}{3} + 2n\pi \quad \text{(general solutions for sine function)}
   \]

2. First, simplify the equations:
   \[
   x + \frac{\pi}{4} = \frac{\pi}{3} + 2n\pi \quad \text{or} \quad x + \frac{\pi}{4} = \frac{2\pi}{3} + 2n\pi
   \]

3. Solving for \(x\) in both cases:
   \[
   x = \frac{\pi}{3} - \frac{\pi}{4} + 2n\pi = \frac{4\pi - 3\pi}{12} + 2n\pi = \frac{\pi}{12} + 2n\pi
   \]
   \[
   x = \frac{2\pi}{3} - \frac{\pi}{4} + 2n\pi = \frac{8\pi - 3\pi}{12} + 2n\pi = \frac{5\pi}{12} + 2n\pi
   \]

Thus, the general solutions for \(x\) are:
\[
x = \frac{\pi}{12} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{12} + 2n\pi
\]
where \(n\) is any integer.

Would you like further clarification or details on any step?

Here are some related questions you might find interesting:

1. How do you determine the general solution for trigonometric equations?
2. Can you explain why the sine function has two solutions within one period?
3. How can we solve similar equations for other trigonometric functions like cosine or tangent?
4. What role do periodicity and symmetry play in solving trigonometric equations?
5. How does the unit circle help in understanding the solutions to trigonometric equations?

**Tip**: Always remember to account for the periodic nature of trigonometric functions when solving equations.

Solve the equation: sin(x + π/4) = √3/2

Learn how to solve the trigonometric equation sin(x + π/4) = √3/2 with detailed steps. This solution explores key trigonometric concepts, including the sine function and general solutions for equations. Suitable for high school students in Grades 10-12.

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