trigonometry Find the solutions

**robasc** · December 1st 2008, 02:59 PM

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?

**vincisonfire** · December 1st 2008, 03:10 PM

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.
$*x = arctan(\frac{-3\pm 7}{2})$

**nzmathman** · December 1st 2008, 03:16 PM

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 $\frac{pi}{2}<x<pi$
and $\frac{3pi}{2}<x<2pi$

For $tan x = 2$ , there will be a solution $0<x<\frac{pi}{2}$ and $pi<x<\frac{3pi}{2}$

**robasc** · December 1st 2008, 03:28 PM

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?

**robasc** · December 1st 2008, 04:11 PM

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 $

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