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

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

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

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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

$2sintheta-1=0$

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

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