What is the general solution of trigonometric equation #sec^2theta+1-3tantheta=0#?

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Answer 1 · 2016-10-06T14:35:38

#theta=npi+pi/4#

or #theta=npi+tan^(-1)2#

As #sec^2theta=1+tan^2theta#,

#sec^2theta+1-3tantheta=0# can be written as

#1+tan^2theta+1-3tantheta=0#

or #tan^2theta-3tantheta+2=0#

or #tan^2theta-2tantheta-tantheta+2=0#

or #tantheta(tantheta-2)-1(tantheta-2)=0#

or #(tantheta-1)(tantheta-2)=0#

i.e. either #tantheta-1=0# or #tantheta-2=0#

hence, either #tantheta=1# or #tantheta=2#

i.e. #theta=npi+pi/4#

or #theta=npi+tan^(-1)2# i.e. #theta=npi+tan^(-1)2# (in radians using calculator).