Hello! I've been tackling the question 'Express sin3x+sinx as a product and hence solve 1/2(sin3x+sinx)=sin2x ; x∈R' but I'm stumped - I'm not sure whether I've even approached it correctly. This is what I did: sin(3x+x)=sin3x.cosx+sinx.cos3x inserting this into the second equation, 1/2(sin3x.cosx+sinx.cos3x)=2sinx.cosx Moving cosx over 4sinx=sin3x+3sinx sinx=sin3x sin(2x+x)-sinx=0 sin2xcosx+cos2x.sinx-sinx=0 2sinxcos^2x+(cos^2 x - sin^2 x) sinx - sinx = 0 2sinxcos^2 x + cos^2xsinx - sin^3 x -sinx=0 sinx(2cos^2 x + cos^2 x - sin^2 x - 1) = 0 sinx(3cos^2x - sin^2x -1) =0 BRACKETS: 3cos^2x - sin^2x - 1 = 0 3cos^2x - (1- cos^2x) - 1 = 0 3cos^2x - 1 + cos^2x - 1 = 0 4cos^2x - 2 = 0 cos^2x = 1/2 and sinx = 0 But at this point I'm stumped. The answer says x = n∏/2 , and I would've gone about the second equation by doing: cosx = √(1/2) x = α = cos^-1(√(1/2)) but this is obviously wrong... :S Argh, how am I supposed to get that answer? Thanks in advance.