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
 A: You did three mistakes in your proposed solution.
First, you are allowed to divide both sides of an equation for an expression if the expression is not null. In the second step of your solution you divide both sides by $\cos^2 x$, thus $4\sin^2 x\cdot \cos^2 x=\cos^2 x$ is equivalent to $4\sin^2 x = 1$ only if $\cos^2 x\neq 0$. But note that the values of $x$ such that $\cos^2 x = 0$ are solution of your equation and hev to be included in the answer.
Second, $4\sin^2x = 1$ is equivalent to $2\sin x=1\vee 2\sin x=-1$.
Third, both the equations $4\sin^2 x=1$ and $\cos^2 x=0$ have infinitely many solutions in $\mathbb{R}$. For instance, consider $\cos x=0$. This equation has $x=\frac \pi 2 \vee x=\frac 32 \pi$ as solutions in $[0,2\pi)$, that is an interval of periodicity for $f(x)=\cos x$. As a consequence, by periodicity, $\left\{\frac \pi 2+2k\pi, \frac 32 \pi+2k\pi \,:\, k\in \mathbb{Z}\right\}$ are all the solutions of $\cos x =0$.
A: $(4\sin^2x)(\cos^2x)=\cos^2x$ implies $\cos x=0$ or $4\sin^2x=1$. For $0\le x<360^\circ$, the former is true for $x=90^\circ,270^\circ$ while the latter is true when $\sin x=\pm1/2$, which in turn is true of $30^\circ,150^\circ,210^\circ,330^\circ$.
A: $$4\sin^2 x \cos^2x - \cos^2 x= \cos^2 x(4\sin^2 x -1)=0$$ 
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.
