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#