Chapter 5.4, Problem 28E

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 $x=\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{7\pi }{6},\frac{5\pi }{6},\frac{11\pi }{6}$ .

Given data:

The equation is $\cos 2x+\cos 4x=0$ .

Formula used:

Use the double angle identity for cosine in terms of sine and cosine.

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

Calculation:

Simplify the above equation by using double angle identity for cosine.

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

Find the roots by using quadratic formula.

  $\begin{array}{c}\cos x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-1\pm \sqrt{1+4\left(2\right)}}{2\left(2\right)}\\ =\frac{-1\pm \sqrt{9}}{4}\\ =\frac{1}{2}\text{and}-1\end{array}$

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

  $\begin{array}{c}\cos x=-1\\ x=\left(2n+1\right)\pi \end{array}$

Equate the second factor to zero.

  $\begin{array}{c}\cos x=\frac{1}{2}\\ x=2n\pi \pm \frac{\pi }{3}\end{array}$

However, the angle is $2x$ instead of $x$ . So,

  $\begin{array}{c}x=\frac{\left(2n+1\right)\pi }{2}\\ x=n\pi \pm \frac{\pi }{6}\end{array}$

The roots will be as follows.

  $x=\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{7\pi }{6},\frac{5\pi }{6},\frac{11\pi }{6}$

Draw the graph of the given equation.

  

From the above graph, it is clear that the solutions are $x=\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{7\pi }{6},\frac{5\pi }{6},\frac{11\pi }{6}$ .

Therefore, the solutions are $x=\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{6},\frac{7\pi }{6},\frac{5\pi }{6},\frac{11\pi }{6}$ .