What's wrong $\sin(2\theta) = \cos\theta$

Question

When solving $\sin(2\theta) = \cos\theta$ for $\theta$ for all values in the range $[0,2\pi]$ I only get half of the solutions when I reduce $\sin(2\theta) = \cos\theta$ to $\sin\theta = \frac{1}{2}$.

So before reducing the equation the solutions are $\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$ and after reducing the equation the solutions are only $\frac{\pi}{6}, \frac{5\pi}{6}$

I can see how it reduces the solution set, but I want to know 'why' it happens.

How can you know exactly when you're reducing the solution set of an equation when you're reducing the equation itself to a simpler form, especially when it's a big complicated equation when you can't easily see that you're reducing the solution set as in this case?

Is there some information regarding that or am I asking the wrong question?

Answer 1

$$\sin(2\theta) = \cos \iff 2\sin\theta \cos\theta - \cos \theta = 0 \iff \cos \theta(2 \sin\theta - 1) = 0$$ $$\iff \cos \theta = 0 \text{ or } \sin\theta = \frac 12$$

If you canceled $\cos \theta$ from each side of the equation (which is legitimate only for $\cos \theta \neq 0$), then you need to additionally consider the solution to when $\cos \theta = 0$, otherwise you lose that information by canceling that factor.

Answer 2

If $x=\cos\theta$ and $y=\sin\theta$ then $\sin2\theta = 2xy$ and your equation is:

$$2xy = x$$ or $$(2y-1)x=0$$ That's true when $y=\frac{1}{2}$ and when else?

Answer 3

$$\sin(2\theta)=\cos \theta\iff2\sin\theta\cos\theta=\cos\theta\iff(2\sin\theta=1)\lor\boxed{(\cos\theta=0)}$$