Answer
(a) $\theta=60^{\circ}+360^{\circ}k,\,180^{\circ}k,\,300^{\circ}+360^{\circ}k.\,$
(b) $0^{\circ},60^{\circ},180^{\circ},300^{\circ}.$
Work Step by Step
(a)
$\tan\theta-2\cos\theta\tan\theta=0$
or, $\tan\theta(1-2\cos\theta)=0$
either, $\tan\theta=0, or \,\,\,\cos\theta=\frac{1}{2}$
For, $\tan\theta=0$,
$\theta=\tan^{-1}(0)$,
$\theta=0^{\circ} \,\,or \,\,180^{\circ},$
To find all values of $\theta$, we can add $180^{\circ}k$ where k is any integer, since the period of tan function is $180^{\circ}$.
So, $\theta=0^{\circ}+180^{\circ}k\,\,or \,\,180^{\circ}+180^{\circ}k$
So, general solution is $\theta=180^{\circ}k.$
For, $\cos\theta=\frac{1}{2}$,
$\theta=\cos^{-1}(\frac{1}{2})$,
Since, $\cos\theta$ is positive in both quadrants 1 and 4,
$\theta=60^{\circ} \,\,or \,\,360^{\circ}-60^{\circ}=300^{\circ},$
To find all values of $\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of cos function is $360^{\circ}$.
So, $\theta=60^{\circ}+360^{\circ}k\,\,or \,\,300^{\circ}+360^{\circ}k$
So, overall general solution of $\theta=60^{\circ}+360^{\circ}k,\,180^{\circ}k,\,300^{\circ}+360^{\circ}k.\,$
(b)
Now, if $0^{\circ}\leq\theta\lt360^{\circ}$, we get the solutions by substituting k=0, 1.
So, the required values of $\theta$ are $0^{\circ},60^{\circ},180^{\circ},300^{\circ}.$
