Trigonometric inverse functions

  • #1
25
0
i want the most general solution for
sin6x=sin4x-sin2x
 
  • #2
sin6x=sin4x-sin2x

sin6x-sin4x+sin2x=0

Then remember that

[tex]sinP+sinQ=2sin(\frac{P+Q}{2})cos(\frac{P-Q}{2})[/tex]
 
  • #3
thanks
 

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