Chapter 13.5, Problem 16CYU

To determine

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

Answer to Problem 16CYU

  $\theta =0°,90°$

Explanation of Solution

Given information :

We know that, The given information is ${\sin }^2\theta -\sin \theta =0$

Calculation :

The given equation is

  $\begin{array}{l}\sin \theta (\sin \theta -1)=0\\ \Rightarrow \sin \theta =0or\sin \theta =1\\ \Rightarrow \sin \theta =\sin 0°or\sin \theta =\sin 90°\\ \Rightarrow \theta =0°or\theta =90°\\ \Rightarrow \theta =0°,90°\end{array}$