Chapter 7, Problem 42RE

To determine

To find:The solution of the equation $\cos 2x{\csc }^{2}x=2\cos 2x$ in the interval $\left[0,2\pi \right)$ .

Answer to Problem 42RE

Thesolutions of the equation $\cos 2x{\csc }^{2}x=2\cos 2x$ are $x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ .

Explanation of Solution

Given information:

The equation is $\cos 2x{\csc }^{2}x=2\cos 2x$ and interval is $\left[0,2\pi \right)$ .

Calculation:

Simplify the equation.

  $\begin{array}{c}\cos 2x{\csc }^{2}x=2\cos 2x\\ \cos 2x{\csc }^{2}x-2\cos 2x=0\\ \cos 2x\left({\csc }^{2}x-2\right)=0\\ \cos 2x=0\text{or}{\csc }^{2}x=2\end{array}$

Simplify further.

  $\begin{array}{c}2x={\cos }^{-1}\left(0\right)\text{or}{\csc }^{2}x=2\\ 2x={\cos }^{-1}\left(0\right)\text{or}\csc x=\pm \sqrt{2}\\ 2x={\cos }^{-1}\left(0\right)\text{or}x={\csc }^{-1}\left(\pm \sqrt{2}\right)\end{array}$

The solutions for the first equation are $2x=\frac{\pi }{2},\frac{3\pi }{2}$ .

To get all the solutions, add integer multiples of $2\pi$ to these solutions. So, the solutions are of the form, $2x=\frac{\pi }{2}+2k\pi$ and $2x=\frac{3\pi }{2}+2k\pi$ or $x=\frac{\pi }{4}+k\pi$ and $x=\frac{3\pi }{4}+k\pi$ .

The solutions that are in the interval $\left[0,2\pi \right)$ , correspond to $k=0$ and $k=1$ . For all other values of $k$ , the corresponding values of $x$ lie outside this interval. So, the solutions in the interval $\left[0,2\pi \right)$ are $x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ .

In the interval $\left[0,2\pi \right)$ , the solution of the second equation are $x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ .

Therefore, the solutions of the equation $\cos 2x{\csc }^{2}x=2\cos 2x$ are $x=\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ .