Question

Solve the trigonometric equation for all values 0 ≤ 3 < 2π. 3 tan x + √3 = 0

Answer 1

Given the following equation:

`3\tan x+\sqrt[]{3}=0`

first, we can substract the square root of 3 on both sides to get:

`\begin{gathered} 3\tan x+\sqrt[]{3}-\sqrt[]{3}=0-\sqrt[]{3} \\ \Rightarrow3\tan x=-\sqrt[]{3} \end{gathered}`

if we divide by 3 both sides of the equation, we get:

`\begin{gathered} (3\tan x=-\sqrt[]{3})\cdot(1)/(3) \\ \Rightarrow(3)/(3)\tan x=-\frac{\sqrt[]{3}}{3} \\ \Rightarrow\tan x=-\frac{\sqrt[]{3}}{3} \end{gathered}`

using the inverse of the function tangent, we can solve for x:

`\begin{gathered} \tan x=-\frac{\sqrt[]{3}}{3} \\ \Rightarrow x=\tan ^(-1)(-\frac{\sqrt[]{3}}{3})=-(\pi)/(6) \\ \end{gathered}`

since we have that x = -pi/6, but we want the values that are betwen 0 and 2 pi, we have to convert using the following expression:

`x+\pi=-(\pi)/(6)+\pi=(-\pi+6\pi)/(6)=(5\pi)/(6)`

therefore, x = 5/6 pi