In my text book I have this equation: \begin{equation} \cos x + \cos 3x - 1 - \cos 2x = 0 \end{equation}

I tried to solve it for $x$, but I didn't succeed.

This is what I tried: \begin{align} \cos x + \cos 3x - 1 - \cos 2x &= 0 \\ 2\cos 2x \cdot \cos x - 1 - \cos 2x &= 0 \\ \cos 2x \cdot (2\cos x - 1) &= 1 \end{align}

So I clearly didn't choose the right path, since this will only be useful if I become something like $a \cdot b = 0$.

All tips will be greatly appreciated.

Solution (Addition to the accepted answer):

The problem was in writing $\cos 3x$ in therms of $cos x$. Anon pointed out that it was equal to $4\cos^3 x - 3\cos x$ but I had to work may way thru it to actually prove that. So I write it down here, maybe it's of use to anybody else.

\begin{align} \cos 3x &= \cos(2x + x)\\ &= \cos(2x)\cdot \cos x - \sin(2x)\sin x\\ &= (\cos^2 x - \sin^2 x) \cdot \cos x - 2\sin^2 x \cdot \cos^2 x\\ &= \cos^3 x - \sin^2x\cdot \cos x - 2\sin^2x\cdot \cos^2 x\\ &= \cos^3 x - (1 - \cos^2 x)\cos x - 2(1-\cos^2 x)\cos x\\ &= cos^3 x - \cos x + \cos^3 x - 2\cos x + 2\cos^3 x\\ &= 4\cos^3 x - 3\cos x \end{align}

EDIT: Apparently you can write all of the formulas $\cos(n\cdot x)$ with $n \in \{1, 2, 3, …\}$ in terms of $\cos x$. Why didn't my teacher tell me that! I don't have time to proof it myself now, but I'll definitely adapt my answer tomorrow (or any time soon)!


To solve an "equals zero" equation involving polynomials or trigonometric functions, it's generally prudent to keep the $0$ on one side and only work with the other side. Here you could use the double and triple angle formulas (which can be deduced from the addition formulas if need be) given by

$$\cos2x=2\cos^2x-1,\qquad \cos3x=4\cos^3x-3\cos x.$$

Substitute these expressions in and your equation will read $P(\cos x)=0$ for a polynomial $P$. Solve for the roots of the polynomial and then take inverse cosines where allowed. Alternatively, if you're at all familiar with complex analysis you could forego the use of angle formulas and go right to a polynomial by using $\cos\theta=(e^{i\theta}+e^{-i\theta})/2$ and then solving for $e^{i\theta}$ and then $\theta$.

  • $\begingroup$ It was indeed $\cos 3x$ which bothered me. Thank you. However I'm afraid complex analysis is a little bit too difficult for me at this point. Maybe some day... :-) $\endgroup$ – user21385 Jan 31 '12 at 20:24

Hint: Can you write $\cos 2x$ and $\cos 3x$ in terms of $\cos x$?

  • $\begingroup$ I can do $\cos 2x = \cos^2 x - \sin^2 x$ and $\cos(x + y) = \cos x \cdot \cos y - \sin x \cdot \sin y$ (which is the same, I know), but this will always leave me with a sinus. $\endgroup$ – user21385 Jan 31 '12 at 20:12
  • 1
    $\begingroup$ @Ief2: Do you know this: $\cos^2 x + \sin^2 x = 1$? $\endgroup$ – Aryabhata Jan 31 '12 at 20:14
  • 1
    $\begingroup$ @Ief2: Now try $\cos 3x$. $\endgroup$ – Aryabhata Jan 31 '12 at 20:18
  • 4
    $\begingroup$ @Ief2: Not just $\cos 2x$ and $\cos 3x$. $\cos 4x, \cos 5x, \dots, \cos nx$ all can be written in terms of just $\cos x$. Can you try proving it (Hint: Use induction and the formula for $\cos (x+y))$? $\endgroup$ – Aryabhata Jan 31 '12 at 20:27
  • 1
    $\begingroup$ I'm studying for a test for tomorrow, so I don't have the time to do it now. But this intrigues me and I'll definitely will try it tomorrow. I'll come back to your comments. $\endgroup$ – user21385 Jan 31 '12 at 20:37

You can also use the identity:

$$\cos(A)+\cos(B) = 2 \cos(\frac{A+B}{2}) \cos (\frac{A-B}{2})$$


\begin{equation} \begin{split} \cos x + \cos 3x - 1 - \cos 2x &= \cos x + \cos 3x - \cos(0) - \cos 2x \ &= 2\cos(\frac{x+3x}{2})\cos(\frac{x-3x}{2})- 2\cos(\frac{0+2x}{2})\cos(\frac{0-2x}{2}) \end{split} \end{equation}

Then $2 \cos(x)$ becomes a common factor, and you can use the corresponding formula for difference this time to factor the rest....

  • $\begingroup$ How can I split an equation? :D $\endgroup$ – N. S. Jan 31 '12 at 22:00
  • $\begingroup$ \begin{alignX} \end{alignX} (without the X's) will do the trick, no dollar signs needed $\endgroup$ – user21385 Feb 5 '12 at 12:55

Your Answer

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