Chapter 7, Problem 104RE

To determine

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

Answer to Problem 104RE

  $\theta \in \frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}$ .

Explanation of Solution

Given:

The given expression is $\sin \left(3\theta \right)=1$ .

Calculation:

As the given equationis simplified as $\sin \left(3\theta \right)=1$ .

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

Therefore, $3\theta =\left(2n+\frac{1}{2}\right)\pi wheren=0,1,\mathrm{2.....}$

For $n=0$

  $\begin{array}{l}3\theta =\left(2\times 0+\frac{1}{2}\right)\pi =\frac{\pi }{2}\\ \therefore \theta =\frac{\pi }{6}\end{array}$

For $n=1$

  $\begin{array}{l}3\theta =\left(2\times 1+\frac{1}{2}\right)\pi =\frac{5\pi }{2}\\ \therefore \theta =\frac{5\pi }{6}\end{array}$

For $n=2$

  $\begin{array}{l}3\theta =\left(2\times 2+\frac{1}{2}\right)\pi =\frac{9\pi }{2}\\ \therefore \theta =\frac{3\pi }{2}\end{array}$

For $n=3$

  $\begin{array}{l}3\theta =\left(2\times 3+\frac{1}{2}\right)\pi =\frac{13\pi }{2}\\ \therefore \theta =\frac{13\pi }{6}\end{array}$

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

Therefore, the solution is $\theta =\frac{13\pi }{6}$ is refused.

Hence, the solution of $\sin \left(3\theta \right)=1$ is $\theta \in \frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}$ .