Chapter 7, Problem 111RE

To determine

To solve:the given equation on the interval $0\le \theta <2\pi$ .

Answer to Problem 111RE

  $\theta \in \frac{\pi }{6},\frac{5\pi }{6}$ .

Explanation of Solution

Given:

The given expression is $\sin \left(2\theta \right)-\cos \theta -2\sin \theta +1=0$ .

Calculation:

As the given equationis simplified as

  $\begin{array}{l}\sin \left(2\theta \right)-\cos \theta -2\sin \theta +1=0\\ 2\sin \theta \cos \theta -\cos \theta -2\sin \theta +1=0\{\sin \left(2\theta \right)=2\sin \theta \cos \theta \}\\ \Rightarrow \cos \theta \left(2\sin \theta -1\right)-2sin\theta +1=0\\ \therefore \cos \theta \left(2\sin \theta -1\right)=2sin\theta -1\\ \therefore \cos \theta =1\\ \therefore \theta =0^{\circ }\end{array}$ .

Also, the givenequation is simplified as

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

Hence, the solution of $\sin \left(2\theta \right)-\cos \theta -2\sin \theta +1=0$ is $\theta \in \frac{\pi }{6},\frac{5\pi }{6}$ .