Show that two sets of solutions are actually identical.

Question

Specific

Context

Solve the equation $ \sin x = \cos 2x $.

Problem

One solution gave the general solution as

• $ x = \pi/6 + 2\pi n/3 $ or

• $ x = -\pi/2 - 2\pi n $ for any integer $n$.

But another solution gave the general solution

• $ x = \pi/6 + 2\pi n $ or

• $ x = 5\pi/6 + 2\pi n $ or

• $ x = 3\pi/2 + 2\pi n $ for any integer $n$.

Show that these two sets of solutions are actually identical.

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General

How does one, conceptually, prove that two sets of solutions are identical?

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Attempt

@user254665 suggest I do the following (see comment under question). I'm not precisely sure what to make of my findings. :P

• $ \pi/6 + 2\pi n/3 = \pi /6 + 2\pi m \implies n = 3m $
• $ \pi/6 + 2\pi n/3 = 5\pi /6 + 2\pi m \implies n = 1 + 3m $
• $ \pi/6 + 2\pi n/3 = 3\pi /2 + 2\pi m \implies n = 2 + 3m $
• $ -\pi /2 - 2\pi n = \pi /6 + 2\pi m \implies 3n = -1 - 3m $
• $ -\pi /2 - 2\pi n = 5\pi /6 + 2\pi m \implies 3n = -2 - 3m $
• $ -\pi /2 - 2\pi n = \pi /2 2\pi m \implies n = -1 - m $

I do notice something about the first 3 bullets: any integer $n$ can be produced by some integer $m$ in 1 of 3 equations.

However, I can't make heads or tails with my findings.

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Disclaimer

I have not taken any set-theory classes. The problem above was given in a trigonometry textbook. In case you need to explain some set-theory, please don't use unfamiliar notation without explaining it.

Answer 1

I think I got it...

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Preparation

In the attempt, I simplified the 6 equations to the following:

• $ n = 3m $
• $ n = 3m + 1 $
• $ n = 3m + 2 $
• $ 3n = -1 - 3m $
• $ 3n = -2 - 3m $
• $ n = -1 - m $

Let us take the first 3 and last equations. The rest are irrelevant.

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Prove: $n \in m$

See the equations:

• $ n = 3m $
• $ n = 3m + 1 $
• $ n = 3m + 2 $

When you plug any integer $m$, you will result in a integer $n$.

• $ n = 3m $ covers ... $-3$, $0$, $3$, $6$ ...
• $ n = 3m + 1 $ covers ... $-2$, $1$, $4$, $7$ ...
• $ n = 3m + 2 $ covers ... $-1$, $2$, $5$, $8$ ...

The combination of these sets results in ... $-3$, $-2$, $-1$, $0$, $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ ...

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Prove: $m \in n$

See the equation $ n = -1 - m $.

It can be rewritten as $ m = -n - 1 $.

It immediately follows that any integer $n$ will result in all integers $n$.

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Answer

Indeed, both sets are identical.

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Disclaimer

I'd like to thank @user254665 for a hint which led me to the solution.

Answer 2

You just have to check that, modulo 2π, the first set of solutions of the first series is $$x=\frac\pi6,\quad \frac{5\pi}6,\quad\frac{9\pi}6=\frac{3\pi}2,$$ so it's the same as the (corrected) second series.

Also, observe the second set of this series is redundant.