Question

In Problems 64-75, solve each equation on the interval$0\le \theta <2\pi$

Short answer

The solution set is$\theta \in \left\{0,\frac{\pi }{6},\frac{5\pi }{6}\right\}$

Step-by-step solution

Given information

Given expression$\sin \left(2\theta \right)-\cos \theta -2\sin \theta +1=0$

Checking the value lie in the specific interval

Calculating, we get

$$\begin{gathered} \sin \left(2\theta \right)-\cos \theta -2\sin \theta +1=0 \\ 2\sin \theta \cos \theta -\cos \theta -2\sin \theta +1=0 \\ 2\sin \theta (\cos \theta -1)-(\cos \theta -1)=0 \\ (\cos \theta -1)(2\sin \theta -1)=0 \\ \cos \theta -1=0 \\ \cos \theta =1 \\ 2\sin \theta -1=0 \\ 2\sin \theta =1 \\ \sin \theta =\frac{1}{2} \\ \theta \in \left\{0,\frac{\pi }{6},\frac{5\pi }{6}\right\} \end{gathered}$$