Chapter 7.5, Problem 25E

(a)

To determine

To find: The solution of the equation $2\sin \frac{\theta }{3}+\sqrt{3}=0$ .

Answer to Problem 25E

The solutions of the equation $2\sin \frac{\theta }{3}+\sqrt{3}=0$ are $\theta =2k\pi$ .

Explanation of Solution

Given information:

The given equation is $2\sin \frac{\theta }{3}+\sqrt{3}=0$ .

Calculation:

Simplify the equation.

  $\begin{array}{c}2\sin \frac{\theta }{3}+\sqrt{3}=0\\ 2\sin \frac{\theta }{2}=-\sqrt{3}\\ \sin \frac{\theta }{2}=-\frac{\sqrt{3}}{2}\end{array}$

As sine has period $2\pi$ . So first find the solutions in $\left[0,2\pi \right)$ .

Therefore, the solution to the equation in $\left[0,2\pi \right)$ are

  $\begin{array}{c}\frac{\theta }{3}={\sin }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\\ =\frac{4\pi }{3},\frac{5\pi }{3}\end{array}$

As sine repeats its values after every $2\pi$ units.

So, the general solution of the equation is given by

  $\begin{array}{c}\frac{\theta }{3}=\frac{4\pi }{3}+2k\pi ,\frac{5\pi }{3}+2k\pi \\ \theta =4\pi +6k\pi ,5\pi +6k\pi \end{array}$

Therefore, the solutions of the equation $2\sin \frac{\theta }{3}+\sqrt{3}=0$ are $\theta =4\pi +6k\pi$ and $\theta =5\pi +6k\pi$ .

(b)

To determine

To find: The solution of the equation $2\sin \frac{\theta }{3}+\sqrt{3}=0$ in $\left[0,2\pi \right)$ .

Answer to Problem 25E

There is no solution possible for the equation $2\sin \frac{\theta }{3}+\sqrt{3}=0$ in $\left[0,2\pi \right)$ .

Explanation of Solution

Given information:

The given equation is $2\sin \frac{\theta }{3}+\sqrt{3}=0$ and the interval is $\left[0,2\pi \right)$ .

Calculation:

According to solution in part(a) the general solution of the equation $2\sin \frac{\theta }{3}+\sqrt{3}=0$ are $\theta =4\pi +6k\pi$ and $\theta =5\pi +6k\pi$ .

Therefore, there is no solution possible for the equation $2\sin \frac{\theta }{3}+\sqrt{3}=0$ in $\left[0,2\pi \right)$ .