Answer
The solutions of the equation are $\theta =\frac{\pi }{6}+2n\pi $ or $\theta =\frac{5\pi }{6}+2n\pi $, where n is any integer.
Work Step by Step
$\begin{align}
& 4sin\theta -1=2\sin \theta \\
& 4sin\theta -2sin\theta =1 \\
& 2\sin \theta =1 \\
& \sin \theta =\frac{1}{2}
\end{align}$
$sin\frac{\pi }{6}=\frac{1}{2}$, and the solutions of $\sin \theta =\frac{1}{2}$ in [0,2 $\pi $ ) are:
$\theta =\frac{\pi }{6}$
$\begin{align}
& \theta =\pi -\frac{\pi }{6} \\
& =\frac{6\pi }{6}-\frac{\pi }{6} \\
& =\frac{5\pi }{6}
\end{align}$
The period of the sine function is $2\pi $, and the solutions are given by:
$\theta =\frac{\pi }{6}+2n\pi $ or $\theta =\frac{5\pi }{6}+2n\pi $, where n is any integer.
