Which values are two of the possible solutions to the equation?

Sine (5 x minus pi) = StartFraction StartRoot 2 EndRoot Over 2 EndFraction, 0 less-than-or-equal-to x less-than 2 pi

We are tasked with solving the equation:

\[
\sin(5x - \pi) = \frac{\sqrt{2}}{2}, \quad 0 \leq x < 2\pi
\]

### Step 1: General solution for \(\sin\theta = \frac{\sqrt{2}}{2}\)

We know that \(\sin\theta = \frac{\sqrt{2}}{2}\) at specific angles. These angles are:

\[
\theta = \frac{\pi}{4} + 2n\pi \quad \text{or} \quad \theta = \frac{3\pi}{4} + 2n\pi \quad \text{for any integer } n.
\]

### Step 2: Apply substitution

Let \( \theta = 5x - \pi \). So, the equation becomes:

\[
\sin(5x - \pi) = \frac{\sqrt{2}}{2}
\]

Thus:

\[
5x - \pi = \frac{\pi}{4} + 2n\pi \quad \text{or} \quad 5x - \pi = \frac{3\pi}{4} + 2n\pi
\]

### Step 3: Solve for \(x\)

#### Case 1: \( 5x - \pi = \frac{\pi}{4} + 2n\pi \)

Solving for \(x\):

\[
5x = \frac{\pi}{4} + \pi + 2n\pi = \frac{5\pi}{4} + 2n\pi
\]
\[
x = \frac{1}{5} \left( \frac{5\pi}{4} + 2n\pi \right)
\]
\[
x = \frac{\pi}{4} + \frac{2n\pi}{5}
\]

#### Case 2: \( 5x - \pi = \frac{3\pi}{4} + 2n\pi \)

Solving for \(x\):

\[
5x = \frac{3\pi}{4} + \pi + 2n\pi = \frac{7\pi}{4} + 2n\pi
\]
\[
x = \frac{1}{5} \left( \frac{7\pi}{4} + 2n\pi \right)
\]
\[
x = \frac{7\pi}{20} + \frac{2n\pi}{5}
\]

### Step 4: Find solutions within the given range

We need to find the values of \(x\) such that \( 0 \leq x < 2\pi \).

#### Case 1: \( x = \frac{\pi}{4} + \frac{2n\pi}{5} \)

- For \(n = 0\): \(x = \frac{\pi}{4}\)
- For \(n = 1\): \(x = \frac{\pi}{4} + \frac{2\pi}{5} = \frac{9\pi}{20}\)
- For \(n = 2\): \(x = \frac{\pi}{4} + \frac{4\pi}{5} = \frac{13\pi}{20}\)

Continue checking for other values, but let's focus on this for now.

#### Case 2: \( x = \frac{7\pi}{20} + \frac{2n\pi}{5} \)

- For \(n = 0\): \(x = \frac{7\pi}{20}\)
- For \(n = 1\): \(x = \frac{7\pi}{20} + \frac{2\pi}{5} = \frac{17\pi}{20}\)

### Final Solutions:

Two possible solutions for \(x\) are:

\[
x = \frac{\pi}{4} \quad \text{and} \quad x = \frac{7\pi}{20}
\]

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

### Related Questions:
1. What are the general solutions for \(\sin x = \frac{\sqrt{2}}{2}\)?
2. How would you solve a cosine equation similar to this?
3. What are the periodic properties of the sine function?
4. How can you convert trigonometric equations into simpler forms?
5. How would the solution differ if the equation involved \(\cos\) instead of \(\sin\)?

**Tip**: Always check the periodicity of trigonometric functions to capture all possible solutions within a given range.

Which values are two of the possible solutions to the equation? Sine (5x - pi) = sqrt(2)/2, 0 ≤ x < 2pi

Learn how to solve the trigonometric equation sin(5x - pi) = sqrt(2)/2 within the interval 0 ≤ x < 2pi. This detailed solution explains the use of trigonometric identities and periodicity. Suitable for high school students in Grades 10-12.

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