Answer
$(a)$
$\displaystyle \theta=\frac{\pi}{3}+2k\pi $ or
$\displaystyle \theta=\frac{2\pi}{3}+2k\pi$
$(b)$
$\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{2\pi}{3}.$
Work Step by Step
The first task is to isolate the trigonometric function on one side:
Add $(\sqrt{3}-2\sin t)$ to both sides...
$4\sin t-2\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 positive in quadrants I and II,
so angles (in radians) within the interval $0\leq t \lt 2\pi $ that satisfy the equation are
$\displaystyle \frac{\pi}{3}$
and
$\pi-\displaystyle \frac{\pi}{3}=\frac{2\pi}{3}$
Finally, to each individual solution, add multiples of $ 2\pi$ to cover all solutions:
$(a)$
$\displaystyle \theta=\frac{\pi}{3}+2k\pi $ or
$\displaystyle \theta=\frac{2\pi}{3}+2k\pi$
$(b)$
The solutions within the interval $0\leq t \lt 2\pi:$
$\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{2\pi}{3}.$
