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Hi I can't figure out how to solve this equation: $$\cos\left(2x\right) + \cos\left(x\right) - 2 = 0$$

I think I'm supposed to rewrite $\cos\left(2x\right)$ into something else and then go from there. I tried a bunch of rewrites but nothing seems to get me anywhere unfortunately.

edit--

due to recommendation here I did an substitution for $u=\cos\left(x\right)$ But now I'm stuck again. Here's what I did

$2u + u -3 = 0$

$2u/2 + u/2 -3/2 = 0$

Here I used the Reduced quadratic equation to obtain the result $u1 = 1, u2 = -3/2$

S4M1R
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2 Answers2

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Since $|\cos t|\le 1$, we have $\cos x=\cos 2x =1$ etc.

Boris Novikov
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Hint: Use the identity $$\cos(2x)=2\cos^2x-1,$$ and make the substitution $u=\cos x.$ What sort of equation do you get?

As a side note, this approach has the benefit of generality, since there's no need to assume that $x$ is real (though I'm sure you're working in the reals). The simplest path, though, assuming that you are working with the reals, is to take the hint that Daniel and Boris recommend.

Added: You seem to have noticed that the substitution gets you a quadratic equation (though it doesn't look like it, based on your formatting). Namely, we get $$2u^2+u-3=0,$$ so $u=1$ or $u=-\frac32,$ as you saw. In context, this means that we have $\cos x=1$ or $\cos x=-\frac32.$ Since (I assume) we're dealing with real values of $x,$ then $\cos x=-\frac32$ is not possible (as Daniel and Boris pointed out). Hence, $\cos x=1.$ For what values of $x$ is this true?

Cameron Buie
  • 102,994