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`

Explanation

`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`

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