Chapter 10.4, Problem 25E

Solution

To find: the general solution of the given equation.

  $x=\frac{\pi }{2}+2n\pi$ or $x=\frac{3\pi }{2}+2n\pi$ or $x=\frac{\pi }{6}+n\pi$ or $x=\frac{7\pi }{6}+n\pi$ , where $n$ is any integer.

Given:

  $\sqrt{3}\cos x\tan x-\cos x=0$

Concept used:

The sine function is positive in I and II quadrants.

The cosine function is positive in I and IV quadrants.

The tangent function is positive in I and III quadrants.

The sine function is periodic with a period of $2\pi$ .

The cosine function is periodic with a period of $2\pi$ .

The tangent function is periodic with a period of $\pi$ .

Calculation:

Consider,

  $\begin{array}{c}\sqrt{3}\cos x\tan x-\cos x=0\\ \cos x\left(\sqrt{3}\tan x-1\right)=0\\ \cos x=0\text{or}\sqrt{3}\tan x-1=0\\ \cos x=0\text{or}\sqrt{3}\tan x=1\\ \cos x=0\text{or}\tan x=\frac{1}{\sqrt{3}}\\ \cos x=0\text{or}\tan x=\frac{\sqrt{3}}{3}\end{array}$

Now, one solution of $\cos x=0$ in the interval $0\le x<2\pi$ is,

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

The other solution in the interval is,

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

Now, one solution of $\tan x=\frac{\sqrt{3}}{3}$ in the interval $0\le x<2\pi$ is,

  $\begin{array}{c}x={\tan }^{-1}\left(\frac{\sqrt{3}}{3}\right)\\ =\frac{\pi }{6}\end{array}$

The other solution in the interval is,

  $\begin{array}{c}x=\pi +\frac{\pi }{6}\\ =\frac{7\pi }{6}\end{array}$

Now since the cosine function is periodic with a period of $2\pi$ , therefore the general solution is,

  $x=\frac{\pi }{2}+2n\pi$ or $x=\frac{3\pi }{2}+2n\pi$ , where $n$ is any integer.

And since the tangent function is periodic with a period of $\pi$ , therefore the general solution is,

  $x=\frac{\pi }{6}+n\pi$ or $x=\frac{7\pi }{6}+n\pi$ , where $n$ is any integer.