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I'm desperately trying to solve $$0=\cos{2x}+\cos x$$

Am I on the right track when I'm this far? $$\cos x(2\cos x+1)=1$$

I don't know where to go from here. What method can I use to further solve the equation? And how do these methods work?

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    $\begingroup$ Yes, you are on the right track. I would expand $\cos 2x = 2\cos^2 x -1$, then put $c = \cos x$ and solve the resulting quadratic equation in $c$. $\endgroup$
    – MJD
    Jul 7, 2014 at 14:59
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    $\begingroup$ You were close but no apples yet! $\endgroup$
    – Chinny84
    Jul 7, 2014 at 15:00

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remember the identity $\cos (2x)=\cos^2 (x)-\sin^2(x)=2\cos ^2 (x)-1$.

You want to solve $2\cos ^2 (x)-1+\cos (x)=0$. Let $t= \cos (x)$.

The question $2t^2+t-1=0$ and is easier to solve. Can you take it from here?

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  • $\begingroup$ Is there an easy way to find the quadratic equation or is it just practice by doing? $\endgroup$
    – user1211030
    Jul 7, 2014 at 15:03
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    $\begingroup$ These questions (and generally trigonometry questions I find) rely heavily on how well you know trigonometric identities. As you can see, if you remember the identity for $\cos(2x)$ this question is easier. My advice would be to practice and memorize (or even better, prove!) trigonometric identities. $\endgroup$
    – Oria Gruber
    Jul 7, 2014 at 15:07
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Note that $\cos{2x}=cos^2{x}-\sin^2{x}=\cos^2{x}-(1-\cos^2{x})=2\cos^2{x}-1$

Then

$$\cos{2x}+\cos{x}=2\cos^2{x}+\cos{x}-1=(\cos{x}+1)(2\cos{x}-1)=0$$

Thus we solve

$$\cos{x}=-1 \text{ and } \cos{x}=\frac{1}{2}$$

Can you get it from here?

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We have

$$0=\cos 2x+\cos x\iff0=2\cos^2x+\cos x-1$$ so let $t=\cos x$ then $$0=2t^2+t-1$$ and the two roots are $$-1\quad;\quad\frac12$$ so $$\cos x=-1\iff x=\pi+2k\pi$$ and $$\cos x=\frac12\iff x=\left(\frac\pi3+2k\pi\right)\lor\left(x=-\frac\pi3+2k\pi\right)$$

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  • $\begingroup$ Very nice flow! $\endgroup$
    – amWhy
    Jul 8, 2014 at 11:40
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In case you don't remember trig identities, there is a geometric way to solve this. Imagine an origin-centered unit circle. $\cos(\theta)$ is the horizontal component of a radius at angle $\theta$ from the $x$ axis. Now $\cos(\theta)=-\cos(2\theta)$ has a geometric interpretation: the horizontal component of a radius at angle at $\theta$ is the negation of the horizontal component of a radius at angle $2\theta$.

There are two solutions: one where $\theta$ is in the first quadrant (with positive cosine) and $2\theta$ is in the second quadrant (with negative cosine), making a symmetry about the positive $y$ axis, and a similar case with symmetry about the negative $y$ axis. You can see that the first case occurs when $\theta$ is $\frac{1}{3}$ of the way to the negative $x$ axis and the second case occurs when $\theta$ is along the negative $x$ axis.

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  • $\begingroup$ Very nicely done! I like geometric solutions that require more intuition rather than memory or knowledge $\endgroup$
    – Oria Gruber
    Jul 7, 2014 at 15:20
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$$\cos2x=-\cos x=\cos(\pi-x)$$

$$\implies2x=2m\pi\pm(\pi-x)$$ where $m$ is any integer

Now consider the signs one by one.

Observe that the set of solutions by '-' sign is a proper subset of the set of solutions by '+' sign

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  • $\begingroup$ @user1211030, Observe that this is susceptible to generalization $$\cos mx+\cos nx=0$$ $\endgroup$
    – lab bhattacharjee
    Jul 7, 2014 at 17:09

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