Chapter 14.5, Problem 51PPS

To determine

To Solve :The given equation is $2{\sin }^2\theta +(\sqrt{2}-1)\sin \theta =\frac{\sqrt{2}}{2}$

Answer to Problem 51PPS

  $\theta =n\pi +{(-1)}^n\frac{\pi }{6}or\theta =n\pi +{(-1)}^n\frac{\pi }{4}$

Explanation of Solution

Given information :

We know that, the given equation is $2{\sin }^2\theta +(\sqrt{2}-1)\sin \theta =\frac{\sqrt{2}}{2}$

Calculation :

The given equation is

  $\begin{array}{l}2{\sin }^2\theta +(\sqrt{2}-1)\sin \theta =\frac{\sqrt{2}}{2}\\ 2\sqrt{2}{\sin }^2\theta +(2-\sqrt{2})\sin \theta =1\\ 2\sqrt{2}{\sin }^2\theta +2\sin \theta -\sqrt{2}\sin \theta -1=0\\ 2\sin \theta (\sqrt{2}\sin \theta +1)-1(\sqrt{2}\sin \theta +1)=0\\ (2\sin \theta -1)(\sqrt{2}\sin \theta +1)=0\\ \sin \theta =\frac{1}{2}or\sin \theta =\frac{1}{\sqrt{2}}\\ \sin \theta =\sin \frac{\pi }{6}or\sin \theta =\sin \frac{\pi }{4}\\ $ \theta =n\pi +{(-1)}^n\frac{\pi }{6}or\theta =n\pi +{(-1)}^n\frac{\pi }{4}$\end{array}$