genral solution (1 Viewer)

[onebytwo]

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 any help

 

[word.]

i got @=n(pi), @=n(pi)+(-1)^n(pi/2), @=n(pi)+(-1)^n(-pi/2)

the answer is @=npi/2

if you take a look closely they're both the same answer...
but in any case you're both wrong
the answer should be x = n.pi +- pi/4

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 ????

these are also the same answer ...