Chapter 7, Problem 99RE

To determine

To solve:the given equation on the interval $0\le \theta <2\pi$ .

Answer to Problem 99RE

  $\theta \in \frac{\pi }{4},\frac{5\pi }{4}$ .

Explanation of Solution

Given:

The given expression is $2cos\theta +\sqrt{2}=0$ .

Calculation:

As the given equationis simplified as

  $\begin{array}{l}2cos\theta =-\sqrt{2}\\ \cos \theta =-\frac{\sqrt{2}}{2}=-\frac{1}{\sqrt{2}}\end{array}$ .

Now, using the property $\cos \theta =x\Rightarrow \theta ={\cos }^{-1}x$ :

Therefore, $\theta ={\cos }^{-1}\left|\left(-\frac{1}{\sqrt{2}}\right)\right|={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi }{4}$

Also, the interval of $0\le \theta <2\pi$

Since, $\cos \theta <0$ , $\theta \in \text{quadrant 2 and }3$

Therefore, in second quadrant,

  $\theta =\pi -\theta =\pi -\frac{\pi }{4}=\frac{3\pi }{4}$

And, in third quadrant,

  $\theta =\pi +\theta =\pi +\frac{\pi }{4}=\frac{5\pi }{4}$

Hence, the solution of $2cos\theta +\sqrt{2}=0$ is $\theta \in \frac{\pi }{4},\frac{5\pi }{4}$ .