Question

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

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

Answer 1

`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`