Question #41db9

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Answer 1 · 2016-10-05T23:40:15

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

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

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

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

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

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

So

when #tantheta=2=tantan^-1(2)=tanalpha#

#=>theta=npi+alpha." where "n inZZ#

when #tantheta=1=tan(pi/4)#

#=>theta=npi+pi/4" where "n in ZZ#