Given $\cos^2 x = 2 \sin x \cos x$, why can't I cancel $\cos x$ to get $\cos x = 2 \sin x$?

Question

If I have a function where I know $\cos^2 x = 2 \sin x \cos x$. Why can I not cross out $\cos x$ on both sides, because I get different values for $\cos x = 2 \sin x$?

Answer 1

The correct step is as follow

$$\cos^2x = 2\sin x\cos x\iff \cos^2x - 2\sin x\cos x=0\iff \cos x(\cos x - 2\sin x)=0$$

and therefore the original equation is equivalent to the following $2$ equations

$$\cos x=0 \quad \lor \quad \cos x - 2\sin x=0$$

As an alternative, rephrasing that, we can also observe that

$$\cos^2x = 2\sin x\cos x$$

is clearly satisfied for $\cos x=0$ which is a solution then for $\cos x\neq0$ we can cancel out and obtain

$$\cos x = 2\sin x$$

Note that the fact is not specifically related to trigonometric function but is a more general fact indeed

$$f(x)\cdot g(x)=f(x)\cdot h(x)$$

by the same argument is equivalent to the following $2$ equations

$$f(x)=0 \quad \lor \quad g(x)=h(x)$$

Answer 2

You may not divide the two members of an equation by $0$. So you can handle the problem with case analysis:

• if $\cos x=0$, the equation holds;

• else if $\cos x\ne0$, the equation can be reduced to $\cos x=2\sin x$.

Now you solve the two cases independently.