How do you solve cos(ø)-2sin(2ø)-cos(3ø)=1-2sin(ø)-cos(2ø)?

Question

ø is theta

The domain is from [0,2pi) not including 2pi

Hint: Get rid of cosines

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Answer 1 · 2016-11-24T02:17:39

$phi={0,pi,pi/2,pi/3,(5pi)/3}$

when $0<=phi<2pi$

$cosphi-2sin2phi-cos3phi=1-2sinphi-cos2phi$

$=>(cosphi-cos3phi)-2sin2phi=(1-cos2phi)-2sinphi$

Using formula
$color(red)(cosC-cosD=2sin((C+D)/2)sin((D-C)/2))$

$=>(2sin2phisinphi)-2sin2phi=2sin^2phi-2sinphi$

$=>2sin2phi(sinphi-1)-2sinphi(sinphi-1)=0$

$=>2(sinphi-1)(sin2phi-sinphi)=0$

$=>2(sinphi-1)(2sinphicosphi-sinphi)=0$

$=>2sinphi(sinphi-1)(2cosphi-1)=0$

Equating each factor except 2 with zero we get

$sinphi=0$

$=>phi=0 and pi$

as $0<=phi<2pi$

when $sinphi-1=0$

$=>sinphi=1=sin(pi/2)$
$=>phi=pi/2$

when $2cosphi-1=0$

$=>cosphi=1/2=cos(pi/3)=cos(2pi-pi/3)$

$=>phi=pi/3 or (5pi)/3$