Chapter 5.5, Problem 11E

To determine

To calculate: The solution of the equation $\sin 2x-\sin x=0$ in the interval $\left[0,2\pi \right)$ .

Answer to Problem 11E

The solution set of the equation $\sin 2x-\sin x=0$ in the interval $\left[0,2\pi \right)$ is $\left\{0,\frac{\pi }{3},\pi ,\frac{5\pi }{3}\right\}$ .

Explanation of Solution

Given information:

The given equation is $\sin 2x-\sin x=0$ .

Formula used:

Use the sum to product formula.

  $\sin u-\sin v=2\cos \left(\frac{u+v}{2}\right)\sin \left(\frac{u-v}{2}\right)$

Calculation:

Substitute the values of sine angles in sum to product formula.

  $\begin{array}{c}\sin 2x-\sin x=2\cos \left(\frac{2x+x}{2}\right)\sin \left(\frac{2x-x}{2}\right)\\ =2\cos \left(\frac{3x}{2}\right)\sin \left(\frac{x}{2}\right)\end{array}$

Further simplify.

  $2\cos \left(\frac{3x}{2}\right)\sin \left(\frac{x}{2}\right)=0$

The solutions in the interval $\left[0,2\pi \right)$ , substitute the factor $\cos \left(\frac{3x}{2}\right)=0$ .

  $\begin{array}{l}\frac{3x}{2}=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2}\\ x=\frac{\pi }{3},\pi ,\frac{5\pi }{3}\end{array}$

The solutions in the interval $\left[0,2\pi \right)$ , substitute the factor $\sin \left(\frac{x}{2}\right)=0$ .

  $\begin{array}{l}x=0\\ \end{array}$

The solution set is

  $\left\{\frac{\pi }{3},\pi \right\}\cup \left\{0\right\}=\left\{0,\frac{\pi }{3},\pi ,\frac{5\pi }{3}\right\}$ .

Verify the solutions using graphical utility:

Draw the graph for the equation $\sin 2x-\sin x=0$ as shown in Figure(1).

  

Figure(1)

From the above, it is clear that $\left\{0,\frac{\pi }{3},\pi ,\frac{5\pi }{3}\right\}$ are the solutions.

Therefore, the solution of the equation $\sin 2x-\sin x=0$ in the interval $\left[0,2\pi \right)$$\left\{0,\frac{\pi }{3},\pi ,\frac{5\pi }{3}\right\}$ .