Chapter 7, Problem 103RE

To determine

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

Answer to Problem 103RE

  $\theta \in 0,\frac{3\pi }{2},\pi ,\frac{5\pi }{2}$ .

Explanation of Solution

Given:

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

Calculation:

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

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

Therefore, $2\theta =n\pi wheren=0,1,\mathrm{2.....}$

For $n=0$

  $\begin{array}{l}2\theta =n\pi =0\\ \therefore \theta =0\end{array}$

For $n=1$

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

For $n=2$

  $\begin{array}{l}2\theta =\left(2\right)\pi =2\pi \\ \therefore \theta =\pi \end{array}$

For $n=3$

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

For $n=4$

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

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

Therefore, the solution is $\theta =2\pi$ is refused.

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