How do you solve #1-sin(theta)=cos2(theta)#?

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Answer 1 · 2018-08-13T14:54:52

#theta={kpi, k inZZ}uu{kpi+(-1)^k *pi/6 ,k inZZ}#

Here ,

#1-sintheta=cos2theta#

#:.1-sintheta=1-2sin^2theta#

#:.2sin^2theta-sintheta=0#

#:.sintheta(2sintheta-1)=0#

#:.sintheta=0 or2sintheta-1=0#

#:.sintheta=0 or sintheta=1/2#

#(i)sintheta=0=>theta=kpi, k inZZ#

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

#=>theta=kpi+(-1)^k *pi/6 ,k inZZ#

Hence,

#theta={kpi, k inZZ}uu{kpi+(-1)^k *pi/6 ,k inZZ}#