## Trigonometry (11th Edition) Clone

The solution set is $$\{0.6089,1.3424,3.7505,4.484\}$$
$$\tan^2x-5\tan x+3=0$$ 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: $$\{0.6089,1.3424,3.7505,4.484\}$$