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!