Solve the equation $\sin ( 2\theta ) =\dfrac{1}{2}$, where $0\leq \theta <2\pi$.

Solution In the interval $\left[ 0,2\pi \right)$, the sine function has the value $\dfrac{1}{2}$ at $\theta =\dfrac{\pi }{6}$ and at $\theta = \dfrac{5\pi }{6}$ as shown in Figure 90. Since the period of the sine function is ${2\pi }$ and the argument in the equation $\sin ( 2\theta ) =\dfrac{1}{2}$ is $2\theta$, we write the general formula for all the solutions. $$\begin{array}{l@{\quad}l@{\quad}l@{\quad}l@{\quad}l} 2\theta =\dfrac{\pi }{6}+2k\pi & {\rm or} & 2\theta =\dfrac{5\pi }{6}+2k\pi, & & \hbox{ where }k \hbox{ is any integer} \\ \theta =\dfrac{\pi }{12}+k\pi & & \theta =\dfrac{5\pi }{12}+k\pi & & \end{array}$$

The solutions of $\sin ( 2\theta ) =\dfrac{1}{2},$ $0\leq \theta <2\pi$, are $\left\{ \dfrac{\pi }{12}, \dfrac{5\pi }{12}, \dfrac{13\pi }{12} , \dfrac{17\pi }{12}\right\} .$