Trig Equations, Degrees to Radians Score: 1/5 Penalty: none Question Watch Video Show Examples Solve the trigonometric equation for all values \( 0 \leq x<2 \pi \). rees to Radians ns (Level 1) \[ \sin x=0 \] Answer Attempt 1 out of a (-) Additional Solution No Solution

To solve the trigonometric equation \( \sin x = 0 \) for all values \( 0 \leq x < 2\pi \), follow these steps: ### Understanding the Sine Function The sine function equals zero at specific points within its period. Specifically: - \( \sin x = 0 \) at \( x = 0 \), \( x = \pi \), and \( x = 2\pi \) within one full period of \( 2\pi \) radians. ### Applying the Interval Constraint Given the interval \( 0 \leq x < 2\pi \): - **\( x = 0 \)** is included because the interval is inclusive of 0. - **\( x = \pi \)** is included. - **\( x = 2\pi \)** is **excluded** because the interval is **less than** \( 2\pi \). ### Solutions Therefore, the solutions to \( \sin x = 0 \) within the interval \( 0 \leq x < 2\pi \) are: \[ x = 0 \quad \text{radians} \] \[ x = \pi \quad \text{radians} \] ### Summary - **Solutions:** \( x = 0 \) and \( x = \pi \) - **Interval:** \( 0 \leq x < 2\pi \) radians These are the points where the sine of \( x \) equals zero within the specified interval.