How do you find all solutions of the equation in the interval #[0,2pi)# given #tan^2theta+tantheta-12=0#?

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Answer 1 · 2018-03-02T12:30:52

Solutions happen to be
#theta=tan^-1(3),pi+tan^-1(3),pi-tan^-1(4),2pi-tan^-1(4)#

Given:

#tan^2theta+tantheta-12=0#
Let t=#tantheta#

#t^2+t-12=0#
Factorising
#t^2+4t-3t-12=0#

#t(t=4)-3(t+4)=0#

#(t-3)(t+4)=0#

#t-3=0#, and #t=-4#
#tantheta =3#
tan theta is positive in first quadrant and third quadrant

#theta=tan^-1(3)#
valid for angles
#theta=tan^-1(3),pi+tan^-1(3)#

#t=-4#
tantheta is negative in second and fourth quadrants

#theta=tan^-1(-4)=-tan^-1(4)#
valid for angles
#theta=pi-tan^-1(4),2pi-tan^-1(4)#

Solutions happen to be
#theta=tan^-1(3),pi+tan^-1(3),pi-tan^-1(4),2pi-tan^-1(4)#