Questions: Find the solution to the equation sin x - 1 = 0 on 0 ≤ x < 2π

Solution Steps

Start with the equation: \[ \sin x - 1 = 0 \] Add 1 to both sides to isolate \(\sin x\): \[ \sin x = 1 \]

We know that \(\sin x = 1\) at: \[ x = \frac{\pi}{2} + 2\pi n, \quad \text{where } n \text{ is an integer} \] Within the interval \(0 \leq x < 2\pi\), the only solution is: \[ x = \frac{\pi}{2} \]

Substitute \(x = \frac{\pi}{2}\) into the original equation: \[ \sin\left(\frac{\pi}{2}\right) - 1 = 1 - 1 = 0 \] The equation holds true, so \(x = \frac{\pi}{2}\) is the correct solution.

Final Answer