# 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)

$
tan^2 x + 3 tan x - 10 = 0

$

Could someone help me with the steps to solve the equation?

2. Let $*u = tan(x)$*
$
u^2 x + 3 u - 10 = 0
$

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

3. Notice this is in quadratic equation form.
We can let $tan x = y$

$y^2 + 3 y - 10 = 0$
$(y + 5)(y - 2) = 0$ (factorising)

$y = -5$ or $y = 2$

Now solve for $tan x = -5$ and $tan x = 2$

Visualising a tan graph we can deduce that:

For $tan x = -5$, there will be a solution $\frac{pi}{2}
and $\frac{3pi}{2}

For $tan x = 2$, there will be a solution $0 and $pi

4. Cool, I never thought about using the quadratic formula to solve. Anyways, this is what I came up with:

$
(tan (x) - 2) 3(tan (x) + 5)
$

$
arctan(2)= 1.1071487

$

and

$

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:

$
arctan(2) + n\pi
$

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