Answer
The solution set is $$\{0.6089,1.3424,3.7505,4.484\}$$
Work Step by Step
$$\tan^2x-5\tan x+3=0$$
over the interval $[0,2\pi)$
Here we can treat the equation as a quadratic equation $au^2+bu+c=0$ in which $u=\tan x$, $a=1$, $b=-5$ and $c=3$.
So to solve this equation, we apply normal operations when solving a quadratic equation.
- Find $\Delta$: $\Delta=b^2-4ac=(-5)^2-4\times1\times3=25-12=13$
- Find $\tan x$: $$\tan x=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{5\pm\sqrt{13}}{2}$$
Now we continue to find $x$, using the inverse function for tangent.
1) For $\tan x=\frac{5+\sqrt{13}}{2}$
$$x=\tan^{-1}\frac{5+\sqrt{13}}{2}\approx1.3424$$
However, do not forget that the period of a tangent function is only $\pi$, which means $\tan x$ would reach the same value every after $x$ runs another round of $\pi$.
That means, over the interval $[0,2\pi)$, there is one more value of $x$ satisfying $\tan x=\frac{5+\sqrt{13}}{2}$, which is $$x=1.3424+\pi\approx4.484$$
So, $$x=\{1.3424,4.484\}$$
2) For $\tan x=\frac{5-\sqrt{13}}{2}$
$$x=\tan^{-1}\frac{5-\sqrt{13}}{2}\approx0.6089$$
However, do not forget that the period of a tangent function is only $\pi$, which means $\tan x$ would reach the same value every after $x$ runs another round of $\pi$.
That means, over the interval $[0,2\pi)$, there is one more value of $x$ satisfying $\tan x=\frac{5-\sqrt{13}}{2}$, which is $$x=0.6089+\pi\approx3.7505$$
So, $$x=\{0.6089,3.7505\}$$
Combining case 1) and 2), we have the solution set:
$$\{0.6089,1.3424,3.7505,4.484\}$$
