Prove that the general solution of `sin"theta"=sin"alpha"` is given by : `theta=npi+(-1)^nalpha,n in Zdot`

Explanation

`sin theta = sin alpha`

`=>sin theta - sinalpha= 0`

`=>2sin((theta-alpha)/2)cos((theta+alpha)/2) = 0`

`=>sin((theta-alpha)/2) = 0 and cos((theta+alpha)/2) = 0`

`=> (theta-alpha)/2 = mpi and (theta+alpha)/2 = (2m+1)pi/2`

`=>theta = 2mpi+alpha and theta = 2mpi+pi-alpha`

So, the general solution will be `theta = npi +(-1)alpha.`