# Thread: trigonometry Find the solutions

1. ## trigonometry Find the solutions

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?

2. Let $\displaystyle *u = tan(x)$*
$\displaystyle u^2 x + 3 u - 10 = 0$
You find $\displaystyle u = tan(x) = \frac{-3\pm\sqrt(9-4\cdot1\cdot-10)}{2} = \frac{-3\pm 7}{2}$ using the quadratic root formula.
$\displaystyle *x = arctan(\frac{-3\pm 7}{2})$

3. 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 $\displaystyle pi<x<\frac{3pi}{2}$

4. 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?

5. I was wrong, there are four solutions to the problem:

1.1071, 4.2487, 1.7682, 4.9098

because:

$\displaystyle arctan(2) + n\pi$
and
$\displaystyle arctan(-5) + n\pi$