Chapter 7, Problem 102RE

To determine

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

Answer to Problem 102RE

  $\theta \in \frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ .

Explanation of Solution

Given:

The given expression is $\cos \left(2\theta \right)=0$ .

Calculation:

As the given equationis $\cos \left(2\theta \right)=0$ .

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

Therefore, $2\theta =\left(n+\frac{1}{2}\right)\pi wheren=0,1,\mathrm{2.....}$

For $n=0$

  $\begin{array}{l}2\theta =\left(0+\frac{1}{2}\right)\pi =\frac{\pi }{2}\\ \therefore \theta =\frac{\pi }{4}\end{array}$

For $n=1$

  $\begin{array}{l}2\theta =\left(1+\frac{1}{2}\right)\pi =\frac{3}{2}\pi \\ \therefore \theta =\frac{3}{4}\pi \end{array}$

For $n=2$

  $\begin{array}{l}2\theta =\left(2+\frac{1}{2}\right)\pi =\frac{5}{2}\pi \\ \therefore \theta =\frac{5}{4}\pi \end{array}$

For $n=3$

  $\begin{array}{l}2\theta =\left(3+\frac{1}{2}\right)\pi =\frac{7}{2}\pi \\ \therefore \theta =\frac{5}{4}\pi \end{array}$

For $n=4$

  $\begin{array}{l}2\theta =\left(4+\frac{1}{2}\right)\pi =\frac{9}{2}\pi \\ \therefore \theta =\frac{9}{4}\pi \end{array}$

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

Therefore, the solution is $\theta =\frac{9}{4}\pi$ is refused.

Hence, the solution of $\sin \left(2\theta \right)+1=0$ is $\theta \in \frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ .