I was doing a simple trig question when it turned out I was missing several answers. I have read somewhere that it is possible to lose information about the signs when dealing with squares and square roots and wondered if something similar happened here.
The question is to find all values of x which satisfy $\sin 4x=\sin 2x$, where $x \in [0, \pi]$
This is what I did: $2\sin 2x\cos 2x=2\sin x\cos x$ $2(2\sin x\cos x)(\cos^2x-\sin^2x)=2\sin x\cos x$ $2\cos^2x-2\sin^2x=1$ $2\cos^2x-2(1-\cos^2x)=1$ $4\cos^2x-3=0$ $\cos x={\pm\sqrt3\over2}$ $x={\pi\over6},{5\pi\over6},{7\pi\over6},{11\pi\over6}$ Apparently I am missing $x={0},{\pi\over2},{\pi},{3\pi\over2}, 2\pi$. Does anyone know if I did something in my solution which lead to the loss of over half the answers? Likewise, how could I do it differently to retain those answers?