Chapter 5.5, Problem 15E

To determine

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

Answer to Problem 15E

The solution set of the equation $\cos 2x-\cos x=0$ in the interval $\left[0,2\pi \right)$ is $\left\{0,\frac{2\pi }{3},\frac{4\pi }{3}\right\}$ .

Explanation of Solution

Given information:

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

Formula used:

Use the sum to product formula.

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

Calculation:

Substitute the values of given cosine anglesinto sum to product formula.

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

Further simplify.

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

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

  $x=0,\frac{2\pi }{3},\frac{4\pi }{3}$

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

  $x=0$

The solution set is

  $\left\{0,\frac{2\pi }{3},\frac{4\pi }{3}\right\}\cup \left\{0\right\}=\left\{0,\frac{2\pi }{3},\frac{4\pi }{3}\right\}$ .

Verify the solutions using graphical utility:

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

  

Figure(1)

From the above, it is clear that $\left\{0,\frac{2\pi }{3},\frac{4\pi }{3}\right\}$ are the solutions.

Therefore, the solution of the equation $\cos 2x-\cos x=0$ in the interval $\left[0,2\pi \right)$$\left\{0,\frac{2\pi }{3},\frac{4\pi }{3}\right\}$ .