Answer
(a) $\theta=120^{\circ}+360^{\circ}k, 150^{\circ}+360k, 210^{\circ}+360k, 240^{\circ}+360k,$ where k is any integer.
(b) $\theta=120^{\circ}, 150^{\circ},210^{\circ},240^{\circ}.$
Work Step by Step
(a)
$(2\cos\theta+\sqrt 3)\times(2\cos\theta+1)=0$
Therefore, either $\cos\theta=-\frac{\sqrt 3}{2} or -\frac{1}{2}$
When $\cos\theta=-\frac{\sqrt 3}{2},$
$\theta=\cos^{-1}(-\frac{\sqrt 3}{2}),$
We know that $\cos\theta$ is negative in quadrant 2 and 3.
So, $\theta$ must be $180^{\circ}-30^{\circ}$ or $180^{\circ}+30^{\circ}$
i.e. $\theta$ must be $150^{\circ}$ or $210^{\circ}$.
Since the period of cosine function is $360^{\circ}$, we add $360^{\circ}k$ to get all degree solutions of $\theta$, k being any integer value.
So, $\theta=150^{\circ}+360^{\circ}k$ or, $210^{\circ}+360^{\circ}k$
When $\cos\theta=-\frac{1}{2},$
$\theta=\cos^{-1}(-\frac{1}{2}),$
We know that $\cos\theta$ is negative in quadrant 2 and 3.
So, $\theta$ must be $180^{\circ}-60^{\circ}$ or $180^{\circ}+60^{\circ}$
i.e. $\theta$ must be $120^{\circ}$ or $240^{\circ}$.
Since the period of cosine function is $360^{\circ}$, we add $360^{\circ}k$ to get all degree solutions of $\theta$, k being any integer value.
So, $\theta=120^{\circ}+360^{\circ}k$ or, $240^{\circ}+360^{\circ}k$
So, general solution of $\theta$= $120^{\circ}+360^{\circ}k, 150^{\circ}+360^{\circ}k, 210^{\circ}+360^{\circ}k, 240^{\circ}+360^{\circ}k$ for all integral values of k.
(b)
for $0^{\circ}\leq\theta\lt360^{\circ}$, we can get the solutions by substituting k=0.
So, possible values of $\theta$ for k=0, are $120^{\circ},150^{\circ},210^{\circ},240^{\circ}.$
