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There is an equation $$\sin2\theta=\sin\theta$$ We need to show when the right-hand side is equal to the left-hand side for $[0,2\pi]$.


Let's rewrite it as $$2\sin\theta\cos\theta=\sin\theta$$ Let's divide both sides by $\sin\theta$ (then $\sin\theta \neq 0 \leftrightarrow \theta \notin \{0,\pi,2\pi\}$) $$2\cos\theta=1$$ $$cos\theta=\frac{1}{2}$$ $$\theta\in\left\{\frac{\pi}{3},\frac{5\pi}{3}\right\}$$
Now, let's try something different. $$2\sin\theta\cos\theta=\sin\theta$$ $$2\sin\theta\cos\theta-\sin\theta=0$$ $$\sin\theta(2\cos\theta-1)=0$$ We can have the solution when $\sin\theta=0$. $$\sin\theta=0$$ $$\theta \in \left\{0,\pi,2\pi\right\}$$ And when $2\cos\theta-1=0$. $$2\cos\theta-1=0$$ $$2\cos\theta=1$$ $$\cos\theta=\frac{1}{2}$$ $$\theta \in \left\{\frac{\pi}{3},\frac{5\pi}{3}\right\}$$ Therefore the whole solution set is $$\theta \in \left\{0,\pi,2\pi,\frac{\pi}{3},\frac{5\pi}{3}\right\}$$ This is the correct solution.
Why is this happening? In the first approach, the extra solution given by $\sin\theta=0$ is not only non-appearing but actually banned. Both approaches look valid to me, yet the first one yields less solutions than the second one. Is the first approach invalid in some cases? This is not the only case when this happens, so I'd like to know when I need to use the second approach to solve the equation, so I don't miss any possible solutions.

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  • $\begingroup$ "In the first approach, the extra solution given by sinθ=0 is not only non-appearing but actually banned." Well, that answers your question, doesn't it? $\sin \theta =0$ is a legitimate solution but it is specifically banned when it shouldnt be. So it is not valid. "Is the first approach invalid in some cases?" It is valid to find all solutions where what you are dividing out is not $0$. So if you use method one you must also do a case where you solve that what you are dividing by IS zero. Those will be the missing solutions. $\endgroup$ – fleablood Jun 8 '18 at 16:56
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Whenever you divide both sides of an equation by something, you are assuming that the thing you're dividing by is nonzero, because dividing by $0$ is not valid.

So going from $2 \sin \theta \cos \theta = \sin \theta$ to $2 \cos\theta = 1$ is only valid when $\sin\theta \ne 0$.

In general, all this means is that you need to check the $\sin \theta = 0$ case separately. For example, going from $2 \sin\theta \cos\theta = 1$ to $2 \cos\theta = \frac1{\sin\theta}$ is also only valid when $\sin\theta \ne 0$, but you don't lose any solutions by doing this, because values of $\theta$ for which $\sin\theta=0$ weren't solutions to begin with.

But in this particular case, when $\sin\theta = 0$, the equation $2\sin\theta \cos\theta = \sin\theta$ is satisfied, so it's correct to go from this to $$ 2\cos\theta = 1 \text{ or } \sin\theta = 0. $$

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Wrong:$$ab=ac\Longleftrightarrow b=c$$

Correct:$$ab=ac\Longleftrightarrow\cases{b=c\\\text{or}\\a=0}$$

Therefore, if you want to simplify by $a$, you automatically end up with a proof by exhaustion, with at least two cases.

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You only consider solutions for $\cos\theta=\frac{1}{2}$ in the first case. In fact if you perform such cancellation, you should consider also solutions given by $\sin\theta=0$.

Why? Multiply both sides by zero:$$2\cos\theta=1,\,\,2\cos\theta\,\cdot0=1\,\cdot0$$ Bare in mind $\sin\theta$ could be $0$, you have $$2\cos\theta\sin\theta=\sin\theta$$

Division is valid if $\sin\theta\ne0$; however, the equation holds if $\sin\theta=0$. This is why you've to consider the case: you've assumed $\sin\theta\ne0$ at cancellation, which is correct; but the equation doesn't have this assumption. You have to consider the case when this assumption is waived.

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