Chapter 13.5, Problem 32PPS

To determine

To Solve : The given value is $\sin 2\theta -\cos \theta ,0\le \theta \le 2\pi$

Answer to Problem 32PPS

  $\theta =\frac{\pi }{2},\frac{3\pi }{2}or\theta =\frac{\pi }{6},\frac{5\pi }{6}$

Explanation of Solution

Given information :

We know that, the given equation is $\sin 2\theta -\cos \theta =0;0\le \theta \le 2\pi$

Formula Used:

  $\sin 2\theta =2\sin \theta \cos \theta$

Calculation :

The given equation is

  $\begin{array}{l}\sin 2\theta -\cos \theta =0\\ 2\sin \theta \cos \theta -\cos \theta =0\\ \cos \theta (2\sin \theta -1)=0\\ \cos \theta =0or\sin \theta =\frac{1}{2}\\ \theta =\frac{\pi }{2},\frac{3\pi }{2}or\theta =\frac{\pi }{6}or\frac{5\pi }{6}\end{array}$