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Sagot :
Given the following equation:
[tex]3\tan x+\sqrt[]{3}=0[/tex]first, we can substract the square root of 3 on both sides to get:
[tex]\begin{gathered} 3\tan x+\sqrt[]{3}-\sqrt[]{3}=0-\sqrt[]{3} \\ \Rightarrow3\tan x=-\sqrt[]{3} \end{gathered}[/tex]if we divide by 3 both sides of the equation, we get:
[tex]\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}[/tex]using the inverse of the function tangent, we can solve for x:
[tex]\begin{gathered} \tan x=-\frac{\sqrt[]{3}}{3} \\ \Rightarrow x=\tan ^{-1}(-\frac{\sqrt[]{3}}{3})=-\frac{\pi}{6} \\ \end{gathered}[/tex]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:
[tex]x+\pi=-\frac{\pi}{6}+\pi=\frac{-\pi+6\pi}{6}=\frac{5\pi}{6}[/tex]therefore, x = 5/6 pi
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