Solve: sin^2x = (sinx)(cosx) 0 < x < 2pi Find exact solutions. Okay, so I got 0, pi/4, but just wasn't sure how to do this... I know you bring the sinxcosx to the other side, making it sin^2x - sinxcosx = 0 ...and then I tried factoring out sinx leaving me with sinx(sinx - cosx). That's where I got lost. So sinx is 0 at 0 and pi on the unit circle, so two solutions are 0 and pi, but what on earth do I do with sinx - cosx? That's when I started thinking about possible other identities that could help me out, but each left me with more craziness on my page!