Answer
$(a)$
$t=\displaystyle \frac{\pi}{6}+2k\pi $ or
$ t=\displaystyle \frac{11\pi}{6}+2k\pi$
$(b)$
$\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{11\pi}{6}.$
Work Step by Step
The first task is to isolate the trigonometric function on one side:
Add $(2\sqrt{3}-2\cos t)$ to both sides...
$2\sqrt{3}=6\cos t-2\cos t\qquad $ ... simplify
$ 2\sqrt{3}=4\cos t\quad$ ... divide with $4$
$\displaystyle \frac{\sqrt{3}}{2}=\cos t$
Now, we find a reference angle. From the table of characteristic angles, we know that $\displaystyle \cos t\frac{\pi}{6}=\frac{\sqrt{3}}{2}.$
Next, we know that cosine is positive in quadrants I and IV,
so angles (in radians) within the interval $0\leq t \lt 2\pi $ that satisfy the equation are
$\displaystyle \frac{\pi}{6}$
and
$ 2\pi-\displaystyle \frac{\pi}{6}=\frac{11\pi}{6}$
Finally, to each individual solution, add multiples of $ 2\pi$ to cover all solutions:
$(a)$
$t=\displaystyle \frac{\pi}{6}+2k\pi $ or
$ t=\displaystyle \frac{11\pi}{6}+2k\pi$
$(b)$
The solutions within the interval $0\leq t \lt 2\pi:$
$\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{11\pi}{6}.$
