Solving a Trigonometric Equation

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 \(\left\{ \dfrac{\pi }{12}, \dfrac{5\pi }{12}, \dfrac{13\pi }{12} , \dfrac{17\pi }{12}\right\} .\)