How do you solve #(3tan^2x-1)(tan^2x-3)=0# in the interval 0 to 2pi?

Question

8798 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-16T03:38:44

#x = {pi/6,(7pi)/6,(5pi)/6,(11pi)/6,pi/3,(4pi)/3,(2pi)/3,(5pi)/3}#

#(3tan^2x-1)(tan^2x-3)=0#
Either
#(3tan^2x-1)=0 or (tan^2x-3)=0#

#tanx=+-1/sqrt3#

when

#tanx=1/sqrt3=tan(pi/6)=tan(pi+pi/6)#

So # x= pi/6 or (7pi)/6#

when

#tanx=-1/sqrt3=tan(pi-pi/6)=tan(2pi-pi/6)#

So # x= (5pi)/6 or (11pi)/6#

From #(tan^2x-3)=0# we get

#tanx=+-sqrt3#

when
#tanx=+sqrt3=tan(pi/3)=tan(pi+pi/3)#

So #x=pi/3 or (4pi)/3#

Again when

#tanx=-sqrt3=tan(pi-pi/3)=tan(2pi-pi/3)#

So #x=(2pi)/3 or (5pi)/3#