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:

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

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

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

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:

$x+\pi=-\frac{\pi}{6}+\pi=\frac{-\pi+6\pi}{6}=\frac{5\pi}{6}$

therefore, x = 5/6 pi