Chapter 10.4, Problem 17E

Solution

To Solve: $3\tan x-\sqrt{3}=0$for $0\le x<2\pi$

  $x=\frac{\pi }{6}\text{ or x}=\frac{7\pi }{6}$for $0\le x<2\pi$ .

Given information:

The equation $3\tan x-\sqrt{3}=0$

Concept and Formula Used:

To solve the above expression, use the values of inverse trigonometric ratios on the unit circle. Also, tangent function is positive in the first and third quadrant, so two solutions of the form $x,\pi +x$ .

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

Calculation:

The given equation is $3\tan x-\sqrt{3}=0$

Adding $\sqrt{3}$ both the sides

  $\begin{array}{c}3\tan x-\sqrt{3}+\sqrt{3}=0+\sqrt{3}\\ 3\tan x=\sqrt{3}\end{array}$

Dividing both the sides by 3

  $\begin{array}{c}\frac{3\tan x}{3}=\frac{\sqrt{3}}{3}\\ \tan x=\frac{1}{\sqrt{3}}\end{array}$

Taking inverse of tangent function both the sides

  $\begin{array}{c}{\tan }^{-1}\left(\tan x\right)={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ x={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ x=\frac{\pi }{6},\pi +\frac{\pi }{6}=\frac{7\pi }{6}\end{array}$

Conclusion:

  $x=\frac{\pi }{6}\text{ or x}=\frac{7\pi }{6}$for $0\le x<2\pi$ .