I'm trying to solve $\tan(\theta)=2\sin(\theta)$ on the interval $[0,2π)$, but having trouble identifying what I'm doing wrong
$$\tan(\theta)=2\sin(\theta)$$
Using quotient identity: $$\tan(\theta)= \frac{\sin(\theta)}{\cos(\theta)}$$
$$\frac{\sin(\theta)}{\cos(\theta)}=2\sin(\theta)$$
Divide both sides by $\sin(\theta)$
$$\frac{1}{\cos(\theta)}=2$$
Reciprocal identity: $$\frac{1}{\cos(\theta)}=\sec(\theta)$$
$$\sec(\theta)=2$$
$$\theta =\frac{\pi}{3}, \frac{5\pi}{3}$$
However, I know that I'm missing solutions $0$ and $\pi$.
I have seen a solution elsewhere online that moves everything to the LHS and then uses the zero-product-property to solve:
$$\tan(\theta)=2\sin(\theta)$$
$$\frac{\sin(\theta)}{\cos(\theta)}=2\sin(\theta)$$
$$\frac{\sin(\theta)}{\cos(\theta)}-2\sin(\theta)=0$$
Factor out $\sin(\theta)$
$$\sin(\theta)\left(\frac{1}{\cos(\theta)}-2\right)=0$$
Use zero-product-property
$\sin(\theta)=0$ or $\dfrac{1}{\cos(\theta)}-2=0$
From here the solutions are $\theta =0, \dfrac{\pi}{3}, \pi, \dfrac{5\pi}{3}$
Still, I don't understand why solutions were missing from the first method and how might I avoid such a mistake in the future?
Best Answer
$\tan\theta=2\sin\theta$ has a solution at $\theta=0$ and $\theta=\pi$, which is also when $\sin\theta=0$.
Because, of this, when you divided both sides by $\sin\theta$, you divided by zero, and this is illegal. That is why you missed some solutions.
To avoid this in future, try not to divide if possible but use the other method you posted, move everything to the LHS and use the zero-product property.
If you ever divide both sides by something, check if that thing can be zero.
In your case, check if $\sin\theta=0$ is a solution to your equation before dividing by it. It is, so first you must check when $\sin\theta=0$, which is at $0,\pi$. After considering these solutions and excluding them you may safely divide by $\sin\theta$.