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

Question

I think I have to start by moving everything to one side and set the equation equal to zero. Now I am lost and having a lot of trouble, any help would be very much appreciated!

Answer 1

Let's make it simple and let y=tan(3x), then y^5=9y, so y(y^4-9)=y(y^2-3)(y^2+3)=0.

So y=0, ±sqrt(3) are the real roots.

Now we go back to y=tan(3x)=0, ±sqrt(3).

So x=0; tan(3x)=±sqrt(3). Tan(60)=sqrt(3), so 3x=60 and x=20 degrees. But we know that tangent is negative in QII and QIV. So 3x=180-60=120 and x=40 degrees; 3x=360-60=300 so x=100 degrees.

CHECK: x=0: tan(0)=0 so tan^5(3x)=tan(3x)=0.

x=20: 3x=60; tan(60)=sqrt(3) and tan^5(60)=9sqrt(3)=9tan(60).

x=40: 3x=120; tan(120)=-sqrt(3) and tan^5(120)=-9sqrt(3)=9tan(120).

x=100: 3x=300; tan(300)=-sqrt(3) and tan^5(300)=-9sqrt(3)=9tan(300).

But we can add 360 to 0, 60, 120 and 300, then divide by 3 to find all possible angles, so the solution is:

x=(1/3)(360n+0)=120n; (1/3)(360n+60)=120n+20; 120n+40; 120n+100 where n is an integer.