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I study maths as a hobby. I have this problem:

Find in radians the general solution of:

$$\cos x + \cos 3x + \cos 5x = 0$$

I have said:

$\cos 3x + \cos 5x = 2\cos\left( \frac{1}{2}(3x + 5x)\right)\cos\left( \frac{1}{2} (3x - 5x)\right) = 2\cos 4x\cos x$

$\cos x + 2\cos 4x \cdot \cos (x) = 0$

$(1 + 2\cos 4x)\cos x = 0$

$\cos x = 0$ or $\cos 4x = - \frac{1}{2}$

$x = \frac{\pi}{2}$ or $x = \frac{\pi}{3} $

My book gives the answers as:

$(2n + 1)\frac{\pi}{6}, n\pi \pm \frac{\pi}{3}$

So we agree on the second answer but is the book wrong when it says $(2n + 1)\frac{\pi}{6}$?

I get the answer as $x = n\pi + \frac{\pi}{2}$, which can be written as $(2n +1)\frac{\pi}{2}$

Axel
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Steblo
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  • But $n\pi \pm \frac{\pi}{3}$ does not cover all cases when $\cos 4x = -\frac{1}{2}$. Take $x = \pi/6$ as an example. – Math Lover Dec 28 '20 at 11:27
  • The first answer in the book covers cases of $\cos x = 0$ and remaining cases where $\cos 4x = - \frac{1}{2}$. So I think book is right. – Math Lover Dec 28 '20 at 11:30

5 Answers5

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$$\cos 4x =-1/2 \iff \cos4x +\cos \frac{\pi}{3} =0 \iff 2\cos\left(2x +\frac{\pi}{6} \right)\cos\left(2x-\frac{\pi}{6} \right)=0 \\ \iff \cos\left(2x +\frac{\pi}{6} \right) = 0 \text{ or } \cos\left(2x -\frac{\pi}{6} \right) = 0 \\ \iff 2x +\frac{\pi}{6} = \frac{\pi}{2}+n\pi \text{ or } 2x -\frac{\pi}{6} = \frac{\pi}{2}+n\pi \\ \iff 2x = \frac{\pi}{3}+n\pi \text{ or } 2x = \frac{2\pi}{3}+n\pi\\ \iff x=\frac{\pi}{6} + n\frac{\pi}{2} \text{ or }x= \frac{\pi}{3}+n\frac{\pi}{2} \\ \iff x= (3n+1) \frac{\pi}{6} \text{ or } x= (3n+2) \frac{\pi}{6}$$ And as you said, for $\cos x=0$ you need to take $$x=(2n+1) \frac{\pi}{2} = (6n+3)\frac{\pi}{6}$$

This is, in fact, equivalent to the book’s answer. To see this, observe that the union of $3n+1, 3n+2, ,6n+3$ is all the integers $m$ such that $m\equiv 1,2,3,4,5 \pmod 6$, which is further equivalent to the union of the odd numbers and numbers of the form $6n+2$ and $6n-2$. So, the solution set can be written as $$x=(2n+1)\frac{\pi}{6} \ \text{or} \ (6n\pm 2)\frac{\pi}{6} = (2n+1)\frac{\pi}{6} \ \text{or} \ n\pi \pm \frac{\pi}{3}$$

Vishu
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  • I get lost from the 3rd line on. And your answer doesn't match mine or the book's. – Steblo Dec 28 '20 at 11:55
  • @Steblo What exactly didn’t you understand? It shouldn’t matter if it matches the book or your answer, as long as it is mathematically correct. – Vishu Dec 28 '20 at 11:56
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    Good edit @Axel – Vishu Dec 28 '20 at 12:06
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    @Steblo Tavish's answer and the books are equivalent, when you look at the sets of solution they are the same. I could post an explanation later of why is it true if you want. – Axel Dec 28 '20 at 12:33
  • @Steblo The problem here is that you can have solutions with different forms. Let me explain, to give you an example ${2k+1 | k \in \mathbb{Z} } = {4k+1 | k \in \mathbb{Z}} \cup {4k+3 | k \in \mathbb{Z}} $ it is the same set but the expressions look different. Tell me if my explanation is not clear enough. – Axel Dec 28 '20 at 12:37
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    I have added an explanation for the same. – Vishu Dec 28 '20 at 12:50
  • I follow all your working up to the point “This is in fact...” I cannot follow the reasoning why $(2n + 1)\frac{\pi}{2} = 2n + 1)\frac{\pi}{6} $ I think it’s partly because I haven’t yet studied unions of sets. – Steblo Dec 29 '20 at 14:03
  • @Steblo That equality does not hold. All I’m saying is that the set of solutions is of the form $\frac{m\pi}{6}$ where $m$ can be anything, except a multiple of $6$. – Vishu Dec 29 '20 at 14:05
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Breaking up the multiple angles. $$\cos3x=4\cos^3x-3\cos x$$ $$\cos5x=16\cos^5x-20\cos^3x+5\cos x$$ Plugging everything together: $$16\cos^5x-16\cos^3x+3\cos x=0$$ Dividing by $\cos x$, and $\cos x$ can't be zero. Defining and plugging in we get: $$y=\cos^2x$$ $$16y^2-16y+3=0$$ Can you take it from here?

Axel
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Andrew Kovács
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2

Two methods of attack (that haven't been mentionned) for such issues with angles in arithmetical progression:

a) It's more efficient to group extreme expressions, here $\cos(x)+\cos(5x)=2 \cos(3x)cos(2x)$, then factor $\cos(3x)$, which gives you instantly the answer.

b) or (longer and advisable only for more complicated cases) use complex numbers

$$\Re(e^{ix}+e^{3ix}+e^{i5x})=\Re(e^{ix}(1+e^{2ix}+e^{i4x}))$$

and recognize the sum of an arithmetic progression...

Jean Marie
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Method$\#1:$

Using Prosthaphaeresis Formulas on $\cos x+\cos5x,$

we get $0=\cos3x(1+2\cos2x)$

$\implies x=(2m+1)\dfrac\pi6$ or $2x=\dfrac{2\pi(3m\pm1)}3\iff x=\dfrac{\pi(6m\pm2)}6$ where $m$ is any integer

Method$\#2:$

Using Prosthaphaeresis Formulas on $\cos3x+\cos5x,$

we get $0=\cos x(1+2\cos4x)$

$\implies x=(2n+1)\dfrac\pi2=\dfrac{(6n+3)\pi}6$ or $4x=\dfrac{2\pi(3n\pm1)}3\iff x=\dfrac{\pi(3n\pm1)}6$ where $n$ is any integer

Method$\#3:$

Like How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

multiply both sides by $\sin\dfrac{3x-x}2=\sin x\ne0,$ and use Werner Formulas

$$0\cdot2\sin x=\sin2x+\sin4x-\sin2x+\sin6x-\sin4x\implies\sin6x=0$$

$$\implies x=\dfrac{r\pi}6\text{ where }r \text{ is any integer, not divisible by }6\text{ as }\sin x\ne0\text{(why?)}$$

So, we need to establish that the following three sets are equivalent :

$$\{r; 6\nmid r\}; \{2m+1,6m+2,6m-2\};\{6n+3, 3n+1, 3n-1\}$$

Take modulo $6,$

$$2m+1\equiv1,3,5; 6m+2\equiv2; 6m-2\equiv4\pmod6$$

$$6n+3\equiv3;3n+1\equiv1,4;3n-1\equiv2,5\pmod6$$

lab bhattacharjee
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You can rewrite the equation in this form: $$\cos(3x-2x)+\cos 3x+\cos(3x+2x)=0$$ This is equivalent to: $$2\cos 3x \cos 2x + \cos 3x = \cos 3x (2\cos 2x+1)=0$$ Eventually, we have the equations $\cos 3x=0$ and $\cos 2x=-\frac{1}{2}$. $$3x=\pi n +\frac{\pi}{2} \space \space \text{or} \space \space 2x=2\pi n \pm \frac{2\pi}{3}\space \rightarrow \space x=\frac{\pi}{6}(2n+1)\space \space \text{or} \space \space x=\pi n \pm \frac{\pi}{3}$$

This is just one way to look at it. You reached $\cos x=0$ and $\cos 4x =-\frac{1}{2}$ equations in your solution. In this case, your solution set is: $$x=\frac{\pi}{6}(6n+3) \space \space \text{or} \space \space x=\frac{\pi}{6}(3n\pm1) $$ The difference here is correlated with the horizontal compression of trigonometric functions.

Fatma Keskin
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