Answer
$(a)$
$60+180^{o}k$
$(b)$
$60^{o}$ and $240^{o}.$
Work Step by Step
The first task is to isolate the trigonometric function on one side:
add $1$ to both sides...
$\sqrt{3}\cot\theta=1\qquad $ ... divide with $\sqrt{3}$
$\displaystyle \cot\theta= \frac{1}{\sqrt{3}}\quad\times\frac{\sqrt{3}}{\sqrt{3}}\qquad $... rationalize
$\displaystyle \cot\theta=\frac{\sqrt{3}}{3}$
Now, we find a reference angle. From the table of characteristic angles, we know that $\displaystyle \cot 60^{o}=\frac{\sqrt{3}}{3}.$
Next, we know that cotangent is positive in quadrants I and III,
so angles that satisfy the equation are
$60^{o}$
and
$180^{o}+60^{o}=240^{o}$
Finally, to each individual solution, add multiples of $360^{o}$ to cover all solutions:
$(a)$
$\theta=60^{o}+360^{o}k $ or
$\theta=240^{o}+360^{o}k $
In degrees, we have 60, 240, 420, 600, 780,...
which can be combined and written as $60+180^{o}k$
$(b)$
The solutions within the interval $ 0^{o}\leq\theta \lt 360^{o}:$
$60^{o}$ and $240^{o}.$
