<s>[INST] Solving $4\sin^2{x}\cos^2{x}-\cos^2{x}=0$ 
Does anyone know why these answers are wrong?
  $$4\sin^2{x}\cos^2{x}-\cos^2{x}=0$$

$(4\sin^2x)(\cos^2x)=\cos^2x$
$4\sin^2x=1$
$\sin x=1/2$
$x=30° , 150°$
Thank you
 [/INST] $$4\sin^2 x \cos^2x - \cos^2 x= \cos^2 x(4\sin^2 x -1)=0$$ 
so $\cos^2 x = 0$ or $4\sin^2 x -1 = 0$; this gives $\cos x =0$ or $\sin^2 x = \frac{1}{4}$ or
$$\cos x = 0 \lor \sin x = \frac12 \lor \sin x = -\frac12$$
Now apply standard angles.
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