Question

1. solve each equation for ( x ), where ( 0 leq x leq 2pi ).

a) ( (2sin x - 1)cos x = 0 )
b) ( (sin x + 1)^2 = 0 )

Explanation:

Step1: 拆分方程为两个情况

若乘积为0，则因子为0，即：
$2\sin x - 1 = 0$ 或 $\cos x = 0$

Step2: 求解$2\sin x - 1 = 0$

移项得$\sin x = \frac{1}{2}$，在$0 \leq x \leq 2\pi$内，解为：
$x = \frac{\pi}{6}, \frac{5\pi}{6}$

Step3: 求解$\cos x = 0$

在$0 \leq x \leq 2\pi$内，解为：
$x = \frac{\pi}{2}, \frac{3\pi}{2}$

Step4: 合并所有解

将两个情况的解汇总

Answer:

$x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$