I having trouble solving this algebraically:

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


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!


You also have this fashion:

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

  • $\begingroup$ How does this help OP in counting the number of solutions satisfying the given conditions? $\endgroup$ – 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:

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

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 " $k 2 \pi$").

  • $\begingroup$ See the remark I have added to my answer. $\endgroup$ – 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

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

  • $\begingroup$ 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 ? $\endgroup$ – Jean Marie Apr 29 '19 at 21:24
  • $\begingroup$ Please see dnqxt's solution. $\endgroup$ – KY Tang Apr 29 '19 at 22:09
  • $\begingroup$ 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. $\endgroup$ – 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 $π/6,3π/6,5π/6,7π/6,9π/6,11π/6.$


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



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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.