Use the inverse functions where necessary to find all solutions of the equation in the interval [0, 2pi)
$\displaystyle
tan^2 x + 3 tan x - 10 = 0
$
Could someone help me with the steps to solve the equation?
Notice this is in quadratic equation form.
We can let $\displaystyle tan x = y$
$\displaystyle y^2 + 3 y - 10 = 0$
$\displaystyle (y + 5)(y - 2) = 0$ (factorising)
$\displaystyle y = -5$ or $\displaystyle y = 2$
Now solve for $\displaystyle tan x = -5$ and $\displaystyle tan x = 2$
Visualising a tan graph we can deduce that:
For $\displaystyle tan x = -5$, there will be a solution $\displaystyle \frac{pi}{2}<x<pi$
and $\displaystyle \frac{3pi}{2}<x<2pi$
For $\displaystyle tan x = 2$, there will be a solution $\displaystyle 0<x<\frac{pi}{2}$ and $\displaystyle pi<x<\frac{3pi}{2}$
Cool, I never thought about using the quadratic formula to solve. Anyways, this is what I came up with:
$\displaystyle
(tan (x) - 2) 3(tan (x) + 5)
$
$\displaystyle
arctan(2)= 1.1071487
$
and
$\displaystyle
arctan(-5) = -1.373400767
$
But I cannot use -1.37 since it is outside of the scope of [0, 2pi) correct?