1.

One of the general solutions of `4sinthetasin2thetasin4theta=sin3theta`is`(3n+-1)pi/(12),AAn in Z``(4n+-1)pi/9,AAn in Z``(3n+-1)pi/(12),AAn in Z``(3n+-1)pi/3,AAn in Z`

Answer» `4sinthetasin2thetasin4theta=sin3theta`
`4sintheta*sin2theta*sin4theta-(3sintheta-4sin^3theta)=0`
`sintheta(4sin2theta*sin4theta-3+4sin^2theta)=0`
`2(2sinthetasin2theta)-3+2*(2sin^2theta)=0`
`2(cos2theta-cos6theta)-3+2(1-cos2theta)=0`
`2cos2theta-2cos6theta-3+2-2cos2theta=0`
`-2cos6theta-1=0`
`cos6theta=-1/2`
`cos6theta=cos(2/3pi)`
`6theta=2npipm2/3pi`
`theta=(npi)/3pmpmpi/9`
`=(3npm1)/pi/9`
Option D is correct.


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