Solve each equation, giving all solutions (general solutions), tan^5(3x)=9tan(3x)

Let's make it simple and let ${\displaystyle y=\tan \left(3x\right)}$, then ${\displaystyle y^5=9y}$, so ${\displaystyle y(y^4-9)=y(y^2-3)(y^2+3)=0.}$
So ${\displaystyle y=0,\pm \sqrt{3}}$ are the real roots.
Now we go back to ${\displaystyle y=\tan \left(3x\right)=0,\pm \sqrt{3}.}$
So ${\displaystyle x=0,\tan \left(3x\right)=\pm \sqrt{3}.\tan \left(60\right)=\sqrt{3},so3x=60\text{and}x=20}$ degrees. But we know that tangent is negative in QII and QIV. So ${\displaystyle 3x=180-60=120\text{and}x=40}$ degrees, ${\displaystyle 3x=360-60=300sox=100}$ degrees.
CHECK: x=0: ${\displaystyle \tan \left(0\right)=0so{\tan }^5\left(3x\right)=\tan \left(3x\right)=0.}$
x=20: 3x=60, ${\displaystyle \tan \left(60\right)=\sqrt{3}\text{and}{\tan }^5\left(60\right)=9\sqrt{3}=9\tan \left(60\right).}$
x=40: 3x=120, ${\displaystyle \tan \left(120\right)=-\sqrt{3}\text{and}{\tan }^5\left(120\right)=-9\sqrt{3}=9\tan \left(120\right).}$
x=100: 3x=300, ${\displaystyle \tan \left(300\right)=-\sqrt{3}\text{and}{\tan }^5\left(300\right)=-9\sqrt{3}=9\tan \left(300\right).}$
But we can add 360 to 0, 60, 120 and 300, then divide by 3 to find all possible angles, so the solution is:
${\displaystyle x=\left(\frac{1}{3}\right)(360n+0)=120n,\left(\frac{1}{3}\right)(360n+60)=120n+20,120n+40,120n+100}$ where n is an integer.