An equation is give (express your answer in terms of k, where k is any integer)
$$2 sin 3θ + 1 = 0$$
(a) Find all solutions of the equation.
(b) Find the solutions in the interval $[$$0$, $2π$$)$.
Part A (answer on bottom right of picture):
Need help with part b. How do I find all the solutions in the given interval?
Best Answer
Note that equivalently $\sin 3\theta = -\frac{1}{2}$ is also satisfied for $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$, and full revolutions from these points. We can therefore have that
$$ 3\theta \;\; =\;\; \frac{7\pi}{6} + 2\pi k \hspace{2pc} \text{or} \hspace{2pc} 3\theta \;\; =\;\; \frac{11\pi}{6} + 2\pi k $$
where $k \in \mathbb{Z}$. These equations can be rewritten as
$$ \theta \;\; =\;\; \frac{(7 + 12k)\pi}{18} \hspace{2pc} \text{or} \hspace{2pc} \theta \;\; =\;\; \frac{(11+12k)\pi}{18}. $$
To find which values are contained in $[0,2\pi)$, we see which values of $k$ allow the above quantities to be $\geq 0$ but $<2\pi$. The solution set is found to be
$$ \left \{\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{19\pi}{18}, \frac{23\pi}{18}, \frac{31\pi}{18}, \frac{35\pi}{18} \right \}. $$
These values are obtained by plugging in $k=0,1,2$ into the above equations. You can verify that these are all less than $2\pi$ and the only solutions which are nonnegative.