plz help find the general solution to these equations
sin3@ = sin@
i got @=n(pi), @=n(pi)+(-1)^n(pi/2), @=n(pi)+(-1)^n(-pi/2)
the answer is @=npi/2
???
and sin3x=sin2x, i thought it would be quite simple, like, x=npi/(3-2(-1)^n)
but the answer is just 2npi or 4n/5 _+ pi/5 ????
thanks for...