Chapter 5.4, Problem 27E

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 $x=0,\pi ,\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{3},\frac{4\pi }{3},\frac{2\pi }{3},\frac{5\pi }{3}$ .

Given data:

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

Formula used:

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

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

Use the double angle identity for of sine.

  $\sin 2u=2\sin u\cos u$

Calculation:

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

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

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

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

Equate the second factor to zero.

  $\begin{array}{c}1+2\cos 2x=0\\ \cos 2x=-\frac{1}{2}\\ 2x=\frac{2\pi }{3},\frac{4\pi }{3}\\ x=\frac{\pi }{3},\frac{\pi }{3}\end{array}$

However, the angle is $2x$ instead of $x$ . So, add $\pi$ to the existing angles, and all these $x$ -values will be the solutions since they are all in $\left[0,2\pi \right)$ .

  $\begin{array}{c}x=0+\pi \\ =\pi \\ x=\frac{\pi }{2}+\pi \\ =\frac{3\pi }{2}\end{array}$

Similarly,

  $\begin{array}{c}x=\frac{\pi }{3}+\pi \\ =\frac{4\pi }{3}\\ x=\frac{2\pi }{3}+\pi \\ =\frac{5\pi }{3}\end{array}$

Draw the graph of the given equation.

  

From the above graph, it is clear that the solutions are $x=0,\pi ,\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{3},\frac{4\pi }{3},\frac{2\pi }{3},\frac{5\pi }{3}$ .

Therefore, the solutions are $x=0,\pi ,\frac{\pi }{2},\frac{3\pi }{2},\frac{\pi }{3},\frac{4\pi }{3},\frac{2\pi }{3},\frac{5\pi }{3}$ .