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: Recall the general solution for the sine function:

The sine function reaches its maximum value of 1 at specific angles. The general solution for \( \sin(\theta) = 1 \) is given by:

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

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

Here, we have \( \sin(2x) = 1 \). Using the general solution for \( \sin(\theta) = 1 \), we substitute \( \theta = 2x \) into the equation:

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

Step 3: Solve for \( x \):

Divide both sides of the equation by 2:

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

Conclusion:

The value of \( x \) is given by:

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

This means that \( x \) takes values starting from \( \frac{\pi}{4} \), and every \( \pi \) radians thereafter.