For the equation #sin(3θ) + cos(3θ) = 1 - sin(2θ)# then?

Question

Please note that the question has multiple answers, i.e. one or more options may be correct.

A: tanθ = 1 is possible
B: cosθ = 0 is possible
C: tanθ/2 = -1 is possible
D: cosθ/2 = 0 is possible

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Answer 1 · 2018-02-25T07:46:00

#sin3theta+cos3theta=1-sin2theta#

#=>3sintheta-4sin^3theta+4cos^3theta-3costheta=1-sin2theta#

#=>3(sintheta-costheta)-4(sin^3theta-cos^3theta)-(1-2sinthetacostheta)=0#

#=>3(sintheta-costheta)-4(sintheta-costheta)(sin^2theta+ sinthetacostheta+cos^2theta)-(sintheta-costheta)^2=0#

#=>(sintheta-costheta)(3-4-4sinthetacostheta-sintheta+costheta)=0#

#=>(sintheta-costheta)(costheta-4sinthetacostheta-sintheta-1)=0#

So

#sintheta-costheta=0#

#=>tantheta=1#

Hence Option (A) is possible.

When

#costheta-4sinthetacostheta-sintheta-1=0#

#=>1-costheta+4sinthetacostheta+sintheta=0#
#=>2sin^2(theta/2)+2sin(theta/2)cos(theta/2)costheta+2sin(theta/2)cos(theta/2)=0#

#=>2sin(theta/2)(sin(theta/2)+cos(theta/2)costheta+cos(theta/2))=0#

#=sin(theta/2)=0#