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,

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

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