Question

What are the solutions for `sec^2x+6tanx=6` in the interval [0,2pi) ??

Thank you friends

Answer 1

The final solutions are:

#x=tan^-1(-3+sqrt14),qquadtan^-1(-3+sqrt14)+pi, quadtan^-1(-3-sqrt14)+pi,qquadtan^-1(-3-sqrt14)+2pi#

Explanation:

`sec^2x+6tanx=6`

`tan^2x+1+6tanx=6`

`tan^2x+1+6tanx-6=0`

`tan^2x+6tanx-5=0`

Using the quadratic formula:

`tanx=(-6+-sqrt((-6)^2-(4)(1)(-5)))/(2(1))`

`tanx=(-6+-sqrt(36+20))/2`

`tanx=(-6+-sqrt(56))/2`

`tanx=(-6+-2sqrt(14))/2`

`tanx=-3+-sqrt(14)`

This means that:

`x=tan^-1(-3+sqrt14),quadtan^-1(-3-sqrt14)`

Since the `tan` function is periodic for every `pi` units, we add `pik` to each answer to represent all the possible solutions:

`x=tan^-1(-3+sqrt14)color(blue)+color(blue)(pik),quadtan^-1(-3-sqrt14)color(blue)+color(blue)(pik)`

After using a calculator, you find that the only four answers on the given interval are:

`x=tan^-1(-3+sqrt14),qquadtan^-1(-3+sqrt14)+pi,quadtan^-1(-3-sqrt14)+pi,qquadtan^-1(-3-sqrt14)+2pi`

Those are the answers. Hope this helped!