The given equation is:
[tex]9\tan ^3x\text{ = 3 tanx}[/tex]
Let y = tan x, the equation becomes:
[tex]\begin{gathered} 9y^3=3y \\ 9y^3-3y=0 \\ 3y(3y^2-1)=0 \\ \text{When 3y = 0} \\ y\text{ = }\frac{0}{3} \\ y\text{ = 0} \\ \text{When 3y}^2-1\text{ = 0} \\ 3y^2=1 \\ y^2=\frac{1}{3} \\ y\text{ = }\pm\frac{1}{\sqrt[]{3}} \end{gathered}[/tex]
Since y = tan x:
[tex]\begin{gathered} \tan \text{ x = 0} \\ x=tan^{-1}(0) \\ x\text{ = }k\pi\text{ where k = }\ldots..,\text{ -2, -1, 0, 1, 2}\ldots\ldots \\ x\text{ = }\ldots.-2\pi,\text{ -}\pi,\text{ 0, }\pi,\text{ 2}\pi\ldots. \\ In\text{ the interval \lbrack{}0, 2}\pi)\colon \\ x\text{ = 0, }\pi \end{gathered}[/tex]
Therefore, when tan x = 0, we got two solutions in the interval (0, 2π]. That is, x = 0, π
Also:
[tex]\begin{gathered} \tan \text{ x = }\pm\frac{1}{\sqrt[]{3}} \\ x=tan^{-1}(\pm\frac{1}{\sqrt[]{3}}) \\ In\text{ the interval \lbrack{}0, 2}\pi\rbrack,\text{ x = }\frac{\pi}{6},\text{ }\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6} \end{gathered}[/tex]
Therefore, the solutions to the equation in the interval (0, 2π] are:
