I am learning precalculus, and I understand how to obtain first two solutions, but I don't understand where did last four solutions came from:
$\endgroup$ 2All values of $\theta$ in the interval $[0,2\pi]$ that satisfy $\sin 3\theta=1/2$ are $$\theta = \frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}, \frac{25\pi}{18}, \frac{29\pi}{18}$$
4 Answers
$\begingroup$Write out $3\theta$: $\frac{\pi}{6}$, $\frac{5\pi}{6}$, $\frac{13\pi}{6}$, $\frac{17\pi}{6}$, $\frac{25\pi}{6}$, $\frac{29\pi}{6}$. Now subtract $\frac{\pi}{6}$ from $\frac{13\pi}{6}$, and you get $\frac{12\pi}{6}=2\pi$. The sine function is periodic, with this period. You get all the other by adding $2\pi$ or $4\pi$ to the first solutions.
Maybe a more obvious way of thinking about the problem is to say $\alpha=3\theta$, and find all solutions of $\sin\alpha=\frac 12$ in the interval $[0,6\pi]$.
$\endgroup$ $\begingroup$$$\theta \in \left[0, 2\pi\right]\implies 3\theta \in \left[0, 6\pi\right]$$$$\sin 3\theta = \dfrac{1}{2}$$$$3\theta = \frac{\pi}{6}+2n\pi, \dfrac{5\pi}{6}+2n\pi, n \in \Bbb Z$$But $3\theta \in \left[0, 6\pi\right]$, so $$3\theta = \dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{13\pi}{6},\dfrac{17\pi}{6},\dfrac{25\pi}{6},\dfrac{29\pi}{6}$$Therefore$$\theta = \dfrac{\pi}{18}, \dfrac{5\pi}{18}, \dfrac{13\pi}{18},\dfrac{17\pi}{18},\dfrac{25\pi}{18},\dfrac{29\pi}{18}$$
$\endgroup$ $\begingroup$Consider the diagram below.
When does $\sin\theta = \sin\varphi$?
The sine of a directed angle is the $y$-coordinate of the point where its terminal side intersects the unit circle. Clearly, one solution is when $\theta = \varphi$. By symmetry, $\sin\theta = \sin(\pi - \theta)$, so $\varphi = \pi - \theta$ is another solution. Moreover, any angle coterminal with $\theta$ or $\pi - \theta$ will also have the same sine. Since coterminal angles differ by integer multiples of $2\pi$, $\sin\theta = \sin\varphi$ if $$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$or $$\varphi = \pi - \theta + 2m\pi, m \in \mathbb{Z}$$
Solve the equation $\sin(3\theta) = \dfrac{1}{2}$ in the interval $[0, 2\pi]$.
A particular solution of the equation $$\sin\alpha = \frac{1}{2}$$is $$\alpha = \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$$Substituting $3\theta$ for $\alpha$ yields$$\sin(3\theta) = \sin\left(\frac{\pi}{6}\right)$$Thus, we have the general solution\begin{align*} 3\theta & = \frac{\pi}{6} + 2k\pi, k \in \mathbb{Z} & 3\theta & = \pi - \frac{\pi}{6} + 2m\pi, m \in \mathbb{Z}\\ \theta & = \frac{\pi}{18} + \frac{2k\pi}{3}, k \in \mathbb{Z} & 3\theta & = \frac{5\pi}{6} + 2m\pi, m \in \mathbb{Z}\\ & & \theta & = \frac{5\pi}{18} + \frac{2m\pi}{3}, m \in \mathbb{Z} \end{align*}Since we seek solutions in the interval $[0, 2\pi]$, $k = 0, 1, 2$ and $m = 0, 1, 2$, which yields\begin{align*} \theta & = \frac{\pi}{18}, \frac{13\pi}{18}, \frac{25\pi}{18} & \theta & = \frac{5\pi}{18}, \frac{17\pi}{18}, \frac{29\pi}{18} \end{align*}
$\endgroup$ $\begingroup$$$ 3 \theta = 30^{\circ}, (180-30)^{\circ} $$ to which add multiples of $360^{\circ}$
That is now
$$ \theta = 10^{\circ}, (60-10)^{\circ} $$ to which add multiples of $120^{\circ}.$
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