### Instruction: Find the sum of the roots of $\tan^2x-9\tan x+1=0$ that are between $x=0$ and $x=2\pi$ radians.### Response: I notice that this equation is a quadratic in terms of $\tan x$, so I can try to use the quadratic formula to find the roots.
I write $\tan x = t$ to make the notation simpler.
Then I have $t^2-9t+1=0$.
Using the quadratic formula, I get $t=\frac{9\pm\sqrt{81-4}}{2}$.
Simplifying, I get $t=\frac{9\pm\sqrt{77}}{2}$.
These are the values of $\tan x$ that satisfy the equation, but I need to find the values of $x$.
To do that, I use the inverse tangent function, $\arctan t$.
I get $x=\arctan \frac{9\pm\sqrt{77}}{2}$.
However, these are not the only solutions, because the tangent function is periodic with period $\pi$.
So I need to add multiples of $\pi$ to these values until I cover the interval $[0,2\pi]$.
I start with the smaller value of $t$, which is $\frac{9-\sqrt{77}}{2}$.
I get $x=\arctan \frac{9-\sqrt{77}}{2}$, which is about 0.141 radians.
This is already in the interval, so I don't need to add anything to it.
I also get $x=\arctan \frac{9-\sqrt{77}}{2}+\pi$, which is about 3.283 radians.
This is also in the interval, so I don't need to add anything to it.
I also get $x=\arctan \frac{9-\sqrt{77}}{2}+2\pi$, which is about 6.425 radians.
This is outside the interval, so I stop here.
Next, I look at the larger value of $t$, which is $\frac{9+\sqrt{77}}{2}$.
I get $x=\arctan \frac{9+\sqrt{77}}{2}$, which is about 1.428 radians.
This is in the interval, so I don't need to add anything to it.
I also get $x=\arctan \frac{9+\sqrt{77}}{2}+\pi$, which is about 4.570 radians.
This is also in the interval, so I don't need to add anything to it.
I also get $x=\arctan \frac{9+\sqrt{77}}{2}+2\pi$, which is about 7.712 radians.
This is outside the interval, so I stop here.
Therefore, the sum of the roots of the equation that are between $x=0$ and $x=2\pi$ radians is $0.141+3.283+1.428+4.570=9.422$ radians.