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

Thank you friends

1 Answer
Mar 22, 2018

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!