# How to solve $\cos(2x) + \cos(4x) = 0$ in the interval $[0,2\pi]$ algebraically.

I having trouble solving this algebraically:

Solve on the interval $$[0,2\pi]$$:

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

My problem is that I keep ending up with $$3$$ solutions: $$\pi/2, 3\pi/2$$ and $$\pi/6$$. But when I graphed it on the interval, it showed $$6$$ solutions. I don't understand and I'm feeling stupid 😭

You also have this fashion:

$$\cos{4x}=-\cos{2x}=\cos{(\pi-2x)}\iff 4x=\pi-2x\text{ or } 4x=2x-\pi\bmod[2\pi]$$

• How does this help OP in counting the number of solutions satisfying the given conditions? – Allawonder Apr 30 '19 at 4:08

Set first $$y:=2x$$.

Your equation which is now $$\cos(y)+\cos(2y)=0$$ can be written, using a well-known formula, as:

$$\cos(y)+(2 \cos(y)^2-1)=0$$

$$2Y^2+Y-1=0$$ in variable $$Y:=\cos(y).$$

This equation has discriminant $$\Delta=9$$, thus roots

$$Y_1=-1 \ \text{and} \ Y_2=\tfrac12$$

Thus, we have two equations

$$(a) \ \cos(y)=-1, \ \ \ \text{and} \ \ \ (b) \ \cos(y)=\tfrac12.$$

(a) gives

$$(a') \ y=\pi+k 2 \pi, \text{for any integer} \ k$$

(b) gives

$$(b') \ y=\pm \pi/3+k 2 \pi$$

Remembering that $$y=2x$$, it remains to divide (a') and (b') by $$2$$ to obtain solutions in variable $$x$$ :

The first equation gives, by taking $$k=0$$, then $$k=1$$ :

$$x=\dfrac\pi2, x=\dfrac\pi2+\pi=3\dfrac\pi2$$

For the same reason, (taking $$k=0$$ and $$k=1$$ ; no need to take other values because we would be outside interval $$[0, 2 \pi]$$), the second equation gives

$$x=\dfrac\pi6, x=5\dfrac\pi6, x=7\dfrac\pi6, x=11\dfrac\pi6$$

This indeed makes 6 solutions on interval $$[0,2\pi]$$.

Remark : the solutions you didn't obtained were may be caused by missing the "for any integer $$k$$" (said otherwise : "up to " $$k 2 \pi$$").

• See the remark I have added to my answer. – Jean Marie Apr 29 '19 at 21:33

We have to allow larger value of angle for 2x such that the x values are within the range.

The equation boiled down to

$$cos 2x = \frac{1}{2}$$

Then we have

$$2x = \frac{\pi}{3} \implies x = \frac{\pi}{6}$$ $$2x = \frac{5\pi}{3} \implies x = \frac{5\pi}{6}$$ $$2x = \frac{7\pi}{3} \implies x = \frac{7\pi}{6}$$ $$2x = \frac{11\pi}{3} \implies x = \frac{11\pi}{6}$$

Also $$cos 2x = -1$$, we have

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

• You use twice "The equation boils down to"... Could you say how you have obtained in a logical way your simplified result "$\cos 2x=\dfrac{1}{2}$ for example ? – Jean Marie Apr 29 '19 at 21:24
• Please see dnqxt's solution. – KY Tang Apr 29 '19 at 22:09
• I am sorry but you should say in your text that it comes from solving a quadratic because this is absolutely not evident, for example for somebody who is maybe in high school. – Jean Marie Apr 29 '19 at 22:22

Yet another way is this:

$$\cos(2x)+\cos(4x)=2\cos(3x)\cos x=0,$$ which implies $$\cos(3x)=0,$$ or $$\cos x=0.$$ Thus, $$3x=\frac π2+πk,$$ or $$x=\frac π2+πm,$$ where $$k,m$$ are integers. Finally, since $$x\in[0,2π],$$ we must have $$0\le \frac π3\left(\frac12+k\right)\le2π,$$ which gives $$0\le k\le 5.$$ The second possibility gives $$0\le π\left(\frac12+m\right)\le 2π,$$ which gives $$m=0,1.$$ Thus, we have eight solutions. However, as you can easily verify, the cases $$k=1$$ and $$m=0$$ yield identical roots; also $$k=4$$ and $$m=1$$ yield identical values, so that we have six distinct solutions satisfying all the conditions. For completeness, these are $$π/6,3π/6,5π/6,7π/6,9π/6,11π/6.$$

$$0=\cos4x+\cos2x=2\cos x\cos3x$$

So it sufficient to have $$\cos3x=0$$ as $$\cos3x$$ has a factor $$\cos x$$

Now $$\cos3x=0\implies3x=(2n+1)\dfrac\pi2$$ where $$n$$ is any integer

$$\implies0\le(2n+1)\dfrac\pi2\le6\pi$$

$$\iff0\le2n+1\le12$$

$$2n+1\ge0\iff n\ge-0.5\implies n\ge0$$

Similarly $$2n+1\le12\iff n\le5.5\implies n\le5$$