Score: 1/5 Penalty: none Question Watch Video Show Examples Solve the trigonometric equation for all values \( 0 \leq x<2 \pi \). \[ \sin x=0 \] ons (Level 1) rees to Radlans Answer Attempt 1 out of 2 (-) Additional Solution No Solution Log Out \( x= \) Submit Answer

To solve the trigonometric equation \( \sin x = 0 \) for all values \( 0 \leq x < 2\pi \), we need to determine the angles \( x \) within this interval where the sine function equals zero. ### Step-by-Step Solution: 1. **Understanding the Sine Function:** The sine of an angle is zero at integer multiples of \( \pi \): \[ \sin x = 0 \quad \text{when} \quad x = k\pi \quad \text{for any integer } k \] 2. **Finding Solutions Within the Specified Interval \( [0, 2\pi) \):** We need to find all \( x \) values within \( 0 \leq x < 2\pi \) where \( \sin x = 0 \): \[ \sin x = 0 \quad \Rightarrow \quad x = 0, \pi \] - **At \( x = 0 \):** \( \sin 0 = 0 \) - **At \( x = \pi \):** \( \sin \pi = 0 \) - **At \( x = 2\pi \):** Although \( \sin 2\pi = 0 \), \( 2\pi \) is not included in the interval \( [0, 2\pi) \). 3. **Conclusion:** The solutions within the interval are: \[ x = 0 \quad \text{and} \quad x = \pi \] ### Final Answer: \[ x = 0 \quad \text{and} \quad x = \pi \]