I attempted to solve [tex]sin2x=2sin^2x[/tex] for [tex]0\leq x\leq \pi[/tex] as follows: [tex]2sinxcosx=2sin^2x[/tex] [tex]2sinx(sinx-cosx)=0[/tex] Therefore, [tex]2sinx=0[/tex] (1) [tex]sinx-cosx=0[/tex] (2) (2) -- [tex]cosxtanx-cosx=0[/tex] [tex]cosx(tanx-1)=0[/tex] Therefore, [tex]cosx=0[/tex] (3) [tex]tanx-1=0[/tex] (4) Hence my solutions should be solving equations (1), (3) and (4). i.e. [tex]sinx=0[/tex], [tex]x=0,\pi[/tex] [tex]cosx=0[/tex], [tex]x=\pi/2[/tex] [tex]tanx=1[/tex], [tex]x=\pi/4[/tex] Hence, my solution set is [tex]x=0,\pi/4,\pi/2,\pi[/tex] But testing the solutions (which I wouldn't have done so in test conditions) show that [tex]\pi/2[/tex] does not satisfy the original equation. This means [tex]cosx\neq 0[/tex], but why? I can't figure out where this extraneous solution came from, and which step I made was invalid to cause this. e.g. I never squared or multiplied the equation by anything etc.