Chapter 5.4, Problem 25E

Solution

To find: The solution of the given equation in the interval $\left[0,2\pi \right)$ and use graph to verify the solutions.

The solutions are $x=\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{4},\frac{7\pi }{4}$ .

Given data:

The equation is $\cos x+\cos 3x=0$ .

Formula used:

Use the double angle identity for cosine.

  $\begin{array}{c}\cos 2u=1-2{\sin }^{2}u\\ \cos 2u=2{\cos }^{2}u-1\end{array}$

Use the double angle identity for sine.

  $\sin 2u=2\sin u\cos u$

Calculation:

Simplify the given function by separating $3x$ and using the cosine sum formula.

  $\begin{array}{c}\cos x+\cos \left(2x+x\right)=0\\ \cos x+\cos 2x\cos x-\sin 2x\sin x=0\end{array}$

Simplify the given equation by using double angle identity for sine and cosine.

  $\begin{array}{c}\cos x+\left(1-2{\sin }^{2}x\right)\cos x-2\sin x\cos x\sin x=0\\ \cos x+\cos x-2{\sin }^{2}x\cos x-2{\sin }^{2}x\cos x=0\\ 2\cos x-4{\sin }^{2}x\cos x=0\\ 2\cos x\left(1-2{\sin }^{2}x\right)=0\end{array}$

Simplify further.

  $\cos x\left(2{\cos }^{2}x-1\right)=0$

Use the zero factor property. Equate the first factor to zero.

  $\begin{array}{c}\cos x=0\\ x={\cos }^{-1}\left(0\right)\\ =\frac{\pi }{2},\frac{3\pi }{2}\end{array}$

Equate the second factor to zero.

  $\begin{array}{c}2{\cos }^{2}x-1=0\\ \cos x=\frac{1}{\sqrt{2}}\\ x={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)\\ x=\frac{\pi }{4},\frac{7\pi }{4}\end{array}$

Draw the graph of the given equation.

  

From the above graph, it is clear that the solutions are $x=\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{4},\frac{7\pi }{4}$ .

Therefore, the solutions are $x=\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{4},\frac{7\pi }{4}$ .