0
$\begingroup$

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?

$\endgroup$
3
  • 7
    $\begingroup$ You divided by $\sin x$, which may be zero. Never divide by zero! Its a law. $\endgroup$ – JCAA Jun 28 '20 at 1:05
  • $\begingroup$ Quite trivial, but $2\pi$ also satisfies your equation and is in the indicated interval. $\endgroup$ – Paralyzed_by_Time Jun 28 '20 at 1:07
  • $\begingroup$ @Paralyzed_by_Time That is actually a typo. The interval is actually $[0, 2\pi)$. I'll go fix that right now. $\endgroup$ – Slecker Jun 28 '20 at 1:09
1
$\begingroup$

$\tan\theta=2\sin\theta$ 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 $\sin\theta$.

$\endgroup$
0
$\begingroup$

To avoid the mistake: when dividing by something, you should make sure that that thing is not zero.


Notice that $\tan x=2\sin x$ is equivalent to $$ \frac{\sin x}{\cos x}=2\sin x $$ which is equivalent to $$ \sin x=0\quad\textrm{or}\quad \cos x = \frac{1}{2}. $$

$\endgroup$
0
$\begingroup$

Notice that as JCAA said in the comment, dividing by $\sin(\theta)$ on both sides may by dividing by $0$ since $\sin(\theta)$ could be $0$, something that shouldn't be done. Instead, from:

$$\frac{\sin(\theta)}{\cos(\theta)}=2\sin(\theta)$$

Rearrange the equation to:

$$2\sin(\theta)\cos(\theta)-\sin(\theta)=0$$ $$\sin(\theta)(2\cos(\theta)-1)=0$$ $$$$ Therefore, solutions come when either $\sin(\theta)$ equals $0$ or $2\cos(\theta)-1$ equals $0$. For the first equation, $sin(\theta)=0$, the answers are $0$ and $\pi$ for $\theta\in[0,2\pi)$. The second equation is equivilant to $\cos(\theta)=\frac{1}{2}$, which has solutions $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ for $\theta\in[0,2\pi)$. We therfore have out 4 solutions:

$$\theta=0,\frac{\pi}{3},\pi,\frac{5\pi}{3}$$

$\endgroup$
0
$\begingroup$

$$\tan(\theta)=2\sin(\theta)$$ $$\frac{\sin\theta}{\cos\theta}-2\sin\theta=0$$ $$\sin\theta\left(\frac{1-2\cos\theta}{\cos\theta}\right)=0$$ $$\tan\theta\left(1-2\cos\theta\right)=0$$ $$\tan\theta=0, \ \ \ 1-2\cos\theta=0$$ $$\theta=k\pi,\ \ \ \cos\theta=\frac{1}{2}$$ $$\theta=k\pi,\ \ \ \theta=2k\pi\pm\frac{\pi}{3}$$ Where, $k$ is any integer i.e. $k=0, \pm1, \pm2, \ldots$. For given interval $\theta\in[0, 2\pi)$, substitute $k=0, k=1$ in above general solution to get $$\color{blue}{\theta=0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}}$$

$\endgroup$
0
$\begingroup$

Assuming that $\theta\notin\{\frac{\pi}{2}+k\pi |k\in\mathbb{Z}\}$, we have : $$\tan(\theta) = 2\sin(\theta) \Leftrightarrow \frac{\sin(\theta)}{\cos(\theta)} = 2\sin(\theta) \Leftrightarrow\sin(\theta) = 0 \vee \cos(\theta) = \frac{1}{2}$$
From one side we have : $$\sin(\theta) = 0 \Leftrightarrow \theta \in\{k\pi|k\in\mathbb{Z}\}$$ On the other side we have : $$\cos(\theta) = \frac{1}{2}\Leftrightarrow \theta\in\{\frac{\pi}{3} + 2k\pi|k\in\mathbb{Z}\}\cup\{-\frac{\pi}{3} + 2k\pi|k\in\mathbb{Z}\}$$
Note that we have:
$$(\{\frac{\pi}{3} + 2k\pi|k\in\mathbb{Z}\}\cup\{-\frac{\pi}{3} + 2k\pi|k\in\mathbb{Z}\})\cap\{\frac{\pi}{2}+k\pi |k\in\mathbb{Z}\} = \emptyset$$
$$\{k\pi|k\in\mathbb{Z}\}\cap\{\frac{\pi}{2}+k\pi |k\in\mathbb{Z}\} = \emptyset$$
-This step is imortant, because we gotta make sure that the solution we found are inside $\tan$'s domain.
Now we can say that all the solution are in the following set : $$\{k\pi|k\in\mathbb{Z}\}\cup\{\frac{\pi}{3} + 2k\pi|k\in\mathbb{Z}\}\cup\{-\frac{\pi}{3} + 2k\pi|k\in\mathbb{Z}\}$$

$\endgroup$
0
$\begingroup$

When you divide both sides by something, you can tried to move them to one side instead and factorize them like solving a polynomial.

$$\frac{sin\theta}{cos\theta}=2sin\theta$$ $$\frac{sin\theta}{cos\theta}-2sin\theta=0$$ $$sin\theta(\frac{1}{cos\theta}-2)=0$$

Then making each part equals to $0$ would give out all answers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.