Answer
$(a)$
$t=\displaystyle \frac{4\pi}{3}+2k\pi $ or
$ t=\displaystyle \frac{5\pi}{3}+2k\pi$
$(b)$
$\displaystyle \frac{4\pi}{3}$ and $\displaystyle \frac{5\pi}{3}.$
Work Step by Step
The first task is to isolate the trigonometric function on one side:
Add $(-\sqrt{3}-3\sin t)$ both sides...
$5\sin t-3\sin t=-\sqrt{3}\qquad $ ... simplify
$ 2\sin t=-\sqrt{3}\quad$ ... divide with $2$
$\displaystyle \sin t=-\frac{\sqrt{3}}{2}$
Now, we find a reference angle. From the table of characteristic angles, we know that $\displaystyle \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}.$
Next, we know that sine is negative in quadrants III and IV,
so angles (in radians) within the interval $0\leq t \lt 2\pi: $that satisfy the equation are
$\displaystyle \pi+\frac{\pi}{3}=\frac{4\pi}{3}$
and
$ 2\pi-\displaystyle \frac{\pi}{3}=\frac{5\pi}{3}$
Finally, to each individual solution, add multiples of $ 2\pi$ to cover all solutions:
$(a)$
$t=\displaystyle \frac{4\pi}{3}+2k\pi $ or
$ t=\displaystyle \frac{5\pi}{3}+2k\pi$
$(b)$
The solutions within the interval $0\leq t \lt 2\pi:$
$\displaystyle \frac{4\pi}{3}$ and $\displaystyle \frac{5\pi}{3}.$
