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

Question

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

Please help!

Answer 1

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

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

Answer 2

Set first $y:=2x$.

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

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

which is a quadratic

$2Y^2+Y-1=0$ 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 " $k 2 \pi$").

Answer 3

You also have this fashion:

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

Answer 4

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 $π/6,3π/6,5π/6,7π/6,9π/6,11π/6.$

Answer 5

Using http://mathworld.wolfram.com/WernerFormulas.html,

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

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