Chapter 5.5, Problem 63E

To determine

Find all the solution of the equation in the given interval.

Answer to Problem 63E

The solutions of the given equation in the interval $\left[0,2\pi \right)$ are:-

  $x=\frac{\pi }{6},\frac{5\pi }{6}$

Explanation of Solution

Given:

  $\frac{\cos 2x}{\sin 3x-\sin x}-1=0$

Using reciprocal formula:

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

By setting the factorcsc x equal to zero, find the solutions in the interval $\left[0,2\pi \right)$ -:

  $\begin{array}{cc}\sin x& =\frac{1}{2}\\ x& =\frac{\pi }{6},\frac{5\pi }{6}\end{array}$

Hence the solutions of the given equation in the interval $\left[0,2\pi \right)$ are:-

  $x=\frac{\pi }{6},\frac{5\pi }{6}$

Let's use agraphing utility to graph $\frac{\cos 2x}{\sin 3x-\sin x}-1=0\text{for}0\le x<2\pi$ as follows:-

  

The approximate x-intercepts are:-

  $x=0.52\approx \frac{\pi }{6},x=2.59\approx \frac{5\pi }{6}$

Hence, these values are the approximate solutions of $\frac{\cos 2x}{\sin 3x-\sin x}-1=0$ in the interval $0\le x<2\pi$ .