Chapter 14.5, Problem 31PPS

To determine

To solve :The value of $2{\sin }^2\theta =1,90°<\theta <270°.$

Answer to Problem 31PPS

  $\theta =\frac{3\pi }{4}or\theta =\frac{5\pi }{4}$

Explanation of Solution

Given information :

We know that, the given expression is $2{\sin }^2\theta =1;90°<\theta <270°.$

Formula Used:

  $a^2-b^2=(a-b)(a+b)$

Calculation :

The given equation is

  $\begin{array}{l}2{\sin }^2\theta =1\\ $ {(\sqrt{2}\sin \theta )}^2-1^2=0$$ {(\sqrt{2}\sin \theta )}^2-1^2=0$(\sqrt{2}\sin \theta -1)(\sqrt{2}\sin \theta +1)=0\\ \sin \theta =\frac{1}{\sqrt{2}}or\sin \theta =\frac{-1}{\sqrt{2}}\\ \theta =\frac{3\pi }{4}or\theta =\frac{5\pi }{4}\end{array}$