Find the sum of the roots of $\tan^2x-9\tan x+1=0$ that are between $x=0$ and $x=2\pi$ radians.
By the quadratic formula,
\[\tan x = \frac{9 \pm \sqrt{77}}{2}.\]Let $r_1 = \frac{9 + \sqrt{77}}{2}$ and $r_2 = \frac{9 - \sqrt{77}}{2}.$  Note that $r_1 r_2 = 1.$

Graphing $y = \tan x,$ we see that $\tan x = r_1$ for two angles in $[0,2 \pi],$ and $\tan x = r_2$ for two angles in $[0,2 \pi].$

[asy]
unitsize(1 cm);

draw(graph(tan,0,1.3),red);
draw(graph(tan,pi - 1.3,1.3 + pi),red);
draw(graph(tan,2*pi - 1.3,2*pi),red);
draw((0,tan(-1.3))--(0,tan(1.3)));
draw((pi/2,tan(-1.3))--(pi/2,tan(1.3)),dashed);
draw((3*pi/2,tan(-1.3))--(3*pi/2,tan(1.3)),dashed);
draw((0,0)--(2*pi,0));
draw((pi,0.2)--(pi,-0.2));
draw((2*pi,0.2)--(2*pi,-0.2));
draw((0,2)--(2*pi,2),blue);
draw((0,1/2)--(2*pi,1/2),blue);

label("$\frac{\pi}{2}$", (pi/2,-0.2), S, UnFill);
label("$\pi$", (pi,-0.2), S);
label("$\frac{3 \pi}{2}$", (3*pi/2,-0.2), S, UnFill);
label("$2 \pi$", (2*pi,-0.2), S);

label("$y = \tan x$", (6.5,-1.5),red);
label("$y = \frac{9 + \sqrt{77}}{2}$", (2*pi,2), E, blue);
label("$y = \frac{9 - \sqrt{77}}{2}$", (2*pi,1/2), E, blue);
[/asy]

Let $\alpha = \arctan r_1,$ and let $\beta = \arctan r_2,$ which are two of the solutions.  Note that
\[\tan \left( \frac{\pi}{2} - \alpha \right) = \frac{\sin (\frac{\pi}{2} - \alpha)}{\cos (\frac{\pi}{2} - \alpha)} = \frac{\cos \alpha}{\sin \alpha} = \frac{1}{\tan \alpha} = \frac{1}{r_1} = r_2.\]It follows that $\beta = \frac{\pi}{2} - \alpha,$ or
\[\alpha + \beta = \frac{\pi}{2}.\]The other two solutions are $\alpha + \pi$ and $\beta + \pi.$  Hence, the sum of all four solutions is
\[\alpha + \beta + \alpha + \pi + \beta + \pi = 2 \alpha + 2 \beta + 2 \pi = \boxed{3 \pi}.\]
3\pi
Level 4
Precalculus
Problem: Find the sum of the roots of $\tan^2x-9\tan x+1=0$ that are between $x=0$ and $x=2\pi$ radians.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the sum of the roots of the equation $\tan^2x - 9\tan x + 1 = 0$, we first recognize that this is a quadratic equation in terms of $\tan x$. We can let $y = \tan x$, transforming the equation into $y^2 - 9y + 1 = 0$. 

Next, we apply the quadratic formula, $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -9$, and $c = 1$. This gives us:

y = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{9 \pm \sqrt{81 - 4}}{2} = \frac{9 \pm \sqrt{77}}{2}.

The roots of the equation are $y_1 = \frac{9 + \sqrt{77}}{2}$ and $y_2 = \frac{9 - \sqrt{77}}{2}$. 

Next, we need to find the corresponding angles $x$ for these values of $y$. The tangent function is periodic with a period of $\pi$, meaning for each root $y_i$, there are two angles in the interval $[0, 2\pi)$: $x_i = \tan^{-1}(y_i)$ and $x_i + \pi$. 

Thus, for each root, we have two angles:
1. For $y_1$: $x_1 = \tan^{-1}\left(\frac{9 + \sqrt{77}}{2}\right)$ and $x_1 + \pi$.
2. For $y_2$: $x_2 = \tan^{-1}\left(\frac{9 - \sqrt{77}}{2}\right)$ and $x_2 + \pi$.

Now, we calculate the sum of these roots. The sum of the angles is:

\text{Sum} = x_1 + (x_1 + \pi) + x_2 + (x_2 + \pi) = 2x_1 + 2x_2 + 2\pi = 2(x_1 + x_2 + \pi).

Next, we find $x_1 + x_2$. By the properties of the tangent function, we know that:

x_1 + x_2 = \tan^{-1}(y_1) + \tan^{-1}(y_2).

Using the identity for the sum of arctangents, we have:

\tan^{-1}(y_1) + \tan^{-1}(y_2) = \tan^{-1}\left(\frac{y_1 + y_2}{1 - y_1y_2}\right).

Calculating $y_1 + y_2$ and $y_1y_2$:

y_1 + y_2 = 9, \quad y_1y_2 = 1.

Thus, we find:

x_1 + x_2 = \tan^{-1}\left(\frac{9}{1 - 1}\right).

Since $1 - y_1y_2 = 0$, this indicates that the sum of the angles approaches $\frac{\pi}{2}$, leading to:

x_1 + x_2 = \frac{\pi}{2}.

Now substituting back into the sum of the roots:

\text{Sum} = 2\left(\frac{\pi}{2} + \pi\right) = 2\left(\frac{3\pi}{2}\right) = 3\pi.

Thus, the final answer is $3\pi$.<|END_OF_LATENT|> The final answer is $3\pi$.