Answer
(a) $\theta=30^{\circ}+360^{\circ}k,60^{\circ}+360^{\circ}k,120^{\circ}+360^{\circ}k,150^{\circ}+360^{\circ}k,$ where k is any integer.
(b) $\theta=30^{\circ},60^{\circ},120^{\circ},150^{\circ}.$
Work Step by Step
(a)
$(2sin\theta-\sqrt 3)\times(2\sin\theta-1)=0$
Therefore, either $\sin\theta=\frac{\sqrt 3}{2} or \frac{1}{2}$
For, $\sin\theta=\frac{\sqrt 3}{2}$,
$\theta=\sin^{-1}(\frac{\sqrt 3}{2})$,
Since, $\sin\theta$ is positive in both quadrants 1 and 2,
$\theta=60^{\circ} \,\,or \,\,180^{\circ}-60^{\circ}=120^{\circ},$
To find all values of $\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of sine function is $360^{\circ}$.
So, $\theta=60^{\circ}+360^{\circ}k\,\,or \,\,120^{\circ}+360^{\circ}k$
For, $\sin\theta=\frac{1}{2}$,
$\theta=\sin^{-1}(\frac{1}{2})$,
Since, $\sin\theta$ is positive in both quadrants 1 and 2,
$\theta=30^{\circ} \,\,or \,\,180^{\circ}-30^{\circ}=150^{\circ},$
To find all values of $\theta$, we can add $360^{\circ}k$ where k is any integer, since the period of sine function is $360^{\circ}$.
So, $\theta=30^{\circ}+360^{\circ}k\,\,or \,\,150^{\circ}+360^{\circ}k$
So, overall general solution of $\theta=30^{\circ}+360^{\circ}k,\,60^{\circ}+360^{\circ}k,\,120^{\circ}+360^{\circ}k,\,150^{\circ}+360^{\circ}k.$
(b)
Now, if $0^{\circ}\leq\theta\lt360^{\circ}$, we get the solutions by substituting k=0.
So, the required values of $\theta$ are $30^{\circ},60^{\circ},120^{\circ},150^{\circ}.$