Answer
$(a)$
$\theta=150^{o}+360^{o}k $ or
$\theta=210^{o}+360^{o}k$
$(b)$
$150^{o}$ and $210^{o}$
Work Step by Step
The first task is to isolate the trigonometric function on one side:
Subtract $\sqrt{3}$ to both sides...
$2\cos\theta=-\sqrt{3}\qquad $ ... divide with 2
$\displaystyle \cos\theta=- \frac{\sqrt{3}}{2}$
Now, we find a reference angle. From the table of characteristic angles, we know that $\displaystyle \cos 30^{o}=\frac{\sqrt{3}}{2}.$
Next, we know that cosine is negative in quadrants I and III,
so angles that satisfy the equation are
$180^{o}-30^{o}=150^{o}$
and
$180^{o}+30^{o}=210^{o}$
Finally, to each individual solution, add multiples of $360^{o}$ to cover all solutions:
$(a)$
$\theta=150^{o}+360^{o}k $ or
$\theta=210^{o}+360^{o}k $
$(b)$
The solutions within the interval $ 0^{o}\leq\theta \lt 360^{o}:$
$150^{o}$ and $210^{o}.$
