Answer
$(a)$
$t=\displaystyle \frac{5\pi}{6}+2k\pi $ or
$ t=\displaystyle \frac{7\pi}{6}+2k\pi$
$(b)$
$\displaystyle \frac{5\pi}{6}$ and $\displaystyle \frac{7\pi}{6}.$
Work Step by Step
The first task is to isolate the trigonometric function on one side:
Add $(-2\sqrt{3}-\cos t)$ to both sides...
$5\cos t-\cos t=-2\sqrt{3}\qquad $ ... simplify
$\cos t=-2\sqrt{3}\quad$ ... divide with $4$
$\displaystyle \cos t=-\frac{\sqrt{3}}{2}$
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 negative in quadrants II and III,
so angles (in radians) within the interval $0\leq t \lt 2\pi: $that satisfy the equation are
$\displaystyle \pi-\frac{\pi}{6}=\frac{5\pi}{6}$
and
$\displaystyle \pi+\frac{\pi}{6}=\frac{7\pi}{6}$
Finally, to each individual solution, add multiples of $ 2\pi$ to cover all solutions:
$(a)$
$t=\displaystyle \frac{5\pi}{6}+2k\pi $ or
$ t=\displaystyle \frac{7\pi}{6}+2k\pi$
$(b)$
The solutions within the interval $0\leq t \lt 2\pi:$
$\displaystyle \frac{5\pi}{6}$ and $\displaystyle \frac{7\pi}{6}.$
