Question

How do you solve `3tan^2x+5tan x-1=0` ?

Answer 1

`x = tan^(-1)(-5/6+-sqrt(37)/6) + npi" "` for any integer `n`

Explanation:

Given:

`3tan^2x+5tan x-1=0`

Let:

`t = tan x`

Then our equation becomes:

`3t^2+5t-1=0`

This is in the form:

`at^2+bt+c = 0`

which has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 5^2-4(3)(-1) = 25+12 = 37`

Since this is positive, the quadratic equation in `t` has real roots, but since it is not a perfect square those roots are irrational.

We can use the quadratic formula to find:

`t = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(t) = (-b+-sqrt(Delta))/(2a)`

`color(white)(t) = (-5+-sqrt(37))/6`

That is:

`tan x = -5/6+-sqrt(37)/6`

Note that `tan x` has period `pi`.

So:

`x = tan^(-1)(-5/6+-sqrt(37)/6) + npi" "` for any integer `n`