Chapter 14.5, Problem 14CYU

To determine

To find :The value of $\theta$ in radians.

Answer to Problem 14CYU

  $\theta =0,\frac{\pi }{3}$

Explanation of Solution

Given information :

We know that, The given information is $\sin \frac{\theta }{2}+\cos \theta =1$

Calculation :

The given equation is

  $\begin{array}{l}\text{sin}\frac{\theta }{\text{2}}+{\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}=1\\ \Rightarrow \sin \frac{\theta }{2}+1-2{\sin }^2\frac{\theta }{2}=1\\ \Rightarrow \sin \frac{\theta }{2}-2{\sin }^2\frac{\theta }{2}=0\\ \Rightarrow \sin \frac{\theta }{2}=0or1-2\sin \frac{\theta }{2}=0\\ \Rightarrow \sin \frac{\theta }{2}=\sin 0°or\sin \frac{\theta }{2}=\frac{1}{2}\\ \Rightarrow \sin \frac{\theta }{2}=\sin 0°or\sin \frac{\theta }{2}=\sin \frac{\pi }{6}\\ \Rightarrow \theta =0and\theta =\frac{\pi }{3}\end{array}$