Question

Find the value of \( x \) if \( \sin(2x) = 1 \).

• A. \( x = \frac{\pi}{2} \)
• B. \( x = \frac{\pi}{4} \)
• C. \( x = \frac{\pi}{6} \)
• D. \( x = \frac{3\pi}{4} \)

Hint:

Remember: \( \sin \theta = 1 \) at \( \theta = \frac{\pi}{2} + 2n\pi \), where \( n \) is an integer. Solve for the principal value first.

Answer 1

The Correct Option is B

Step 1: General solution for the sine function:

The sine function's maximum value of 1 occurs at specific angles. The general solution for \( \sin(\theta) = 1 \) is:

\[ \theta = \frac{\pi}{2} + 2n\pi \quad \text{where } n \in \mathbb{Z} \]

Step 2: Applying the solution to \( \sin(2x) = 1 \):

Given \( \sin(2x) = 1 \). By substituting \( \theta = 2x \) into the general solution:

\[ 2x = \frac{\pi}{2} + 2n\pi \quad \text{where } n \in \mathbb{Z} \]

Step 3: Solving for \( x \):

Dividing both sides by 2 yields:

\[ x = \frac{\pi}{4} + n\pi \quad \text{where } n \in \mathbb{Z} \]

Conclusion:

The solutions for \( x \) are:

\[ x = \frac{\pi}{4} + n\pi \quad \text{where } n \in \mathbb{Z} \]

This indicates that \( x \) takes values beginning with \( \frac{\pi}{4} \) and increments by \( \pi \) radians for each integer \( n \).