How do you find the solution to #tantheta+12=2tantheta+11# if #0<=theta<2pi#?

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Answer 1 · 2018-04-08T15:23:57

#color(blue)(pi/4)#

#color(blue)((5pi)/4)#

First collect like terms:

#12-11=2tan(theta)-tan(theta)#

#tan(theta)=1#

#theta=arctan(tan(theta))=arctan(1)=>theta#

#=pi/4+npi#

#n in ZZ#

This is the general solution, for given interval.

#[0,2pi)#

#n=0#

#color(blue)(pi/4)#

#n=1#

#color(blue)((5pi)/4)#