solve the equation on the interval 0 ≤ θ < 2π. sin (3θ) = - 1

Explanation:

Step1: Recall sine - function values

We know that $\sin x=-1$ when $x = \frac{3\pi}{2}+ 2k\pi$, where $k\in\mathbb{Z}$. So for $\sin(3\theta)=-1$, we set $3\theta=\frac{3\pi}{2}+2k\pi$.

Step2: Solve for $\theta$

Divide both sides of the equation $3\theta=\frac{3\pi}{2}+2k\pi$ by 3. We get $\theta=\frac{\pi}{2}+\frac{2k\pi}{3}$.

Step3: Find values of $\theta$ in the given interval

When $k = 0$, $\theta=\frac{\pi}{2}$. When $k = 1$, $\theta=\frac{\pi}{2}+\frac{2\pi}{3}=\frac{3\pi + 4\pi}{6}=\frac{7\pi}{6}$. When $k = 2$, $\theta=\frac{\pi}{2}+\frac{4\pi}{3}=\frac{3\pi+8\pi}{6}=\frac{11\pi}{6}$. When $k = 3$, $\theta=\frac{\pi}{2}+ 2\pi>\ 2\pi$.

Answer:

$\theta=\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}$