Chapter 13.5, Problem 43PPS

To determine

To find : The solution of $4{\sin }^2\theta =4\sin \theta -1$

Answer to Problem 43PPS

  $\theta =180°n+{(-1)}^{n}30°,n\in \mathbb{Z}$

Explanation of Solution

Given information :

We know that, the given equation is $4{\sin }^2\theta =4\sin \theta -1$

Formula Used:

  ${(a-b)}^2=a^2-2ab+b^2$

Calculation :

The given equation is

  $\begin{array}{l}4{\sin }^2\theta -4\sin \theta +1=0\\ $ {(2\sin \theta -1)}^2=0$\sin \theta =\frac{1}{2}\\ \sin \theta =\sin \frac{\pi }{6}\\ $ \theta =180°n+{(-1)}^{n}30°,n\in \mathbb{Z}$\end{array}$