Trigonometric inverse functions

[lizzie]

i want the most general solution for
sin6x=sin4x-sin2x

 

[rock.freak667]

sin6x=sin4x-sin2x

sin6x-sin4x+sin2x=0

Then remember that

$$sinP+sinQ=2sin(\frac{P+Q}{2})cos(\frac{P-Q}{2})$$

 

[lizzie]

thanks