Question #1560d

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Answer 1 · 2016-12-31T01:31:59

#cos2theta-sintheta=0#

#=>cos2theta=sintheta#

#=>cos2theta=cos(pi/2-theta)#

#=>2theta=2npipm(pi/2-theta)#

when

#2theta=2npi+(pi/2-theta)" where " n in ZZ#

#=>3theta=2npi+pi/2#

#=>theta=(2npi)/3+pi/6=(4n+1)pi/6#

when

#2theta=2npi-(pi/2-theta)" where " n in ZZ#

#=>theta=2npi-pi/2=(4n-1)pi/2#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Alternative

#cos2theta-sintheta=0#

#=>1-2sin^2theta-sintheta=0#

#=>2sin^2theta+sintheta-1=0#

#=>2sin^2theta+2sintheta-sintheta-1=0#

#=>2sintheta(sintheta+1)-(sintheta+1)=0#

#=>(sintheta+1)(2sintheta-1)=0#

So

#sintheta+1=0#

#=>sintheta=-1=sin(-pi/2)#

#theta=npi-(-1)^npi/2" where " n in ZZ#

Again

#2sintheta-1=0#

#=>sintheta=1/2=sin(pi/6)#

#theta=npi+(-1)^npi/6" where " n in ZZ#