Question

I need help in math can you please help me please

Answer 1

The given equation is:

`9\tan ^3x\text{ = 3 tanx}`

Let y = tan x, the equation becomes:

`\begin{gathered} 9y^3=3y \\ 9y^3-3y=0 \\ 3y(3y^2-1)=0 \\ \text{When 3y = 0} \\ y\text{ = }(0)/(3) \\ y\text{ = 0} \\ \text{When 3y}^2-1\text{ = 0} \\ 3y^2=1 \\ y^2=(1)/(3) \\ y\text{ = }\pm\frac{1}{\sqrt[]{3}} \end{gathered}`

Since y = tan x:

`\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}`

Therefore, when tan x = 0, we got two solutions in the interval (0, 2π]. That is, x = 0, π

Also:

`\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 = }(\pi)/(6),\text{ }(5\pi)/(6),(7\pi)/(6),(11\pi)/(6) \end{gathered}`

Therefore, the solutions to the equation in the interval (0, 2π] are: