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# Solving this equation algebraically?

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cos x - cos 2x = 0

So far, I've subbed in 2cos$${^2}$$(x) - 1; cos x - 2cos$${^2}$$x +1 = 0

I'm not sure what I can do next?

Thank you! :)

Feb 22, 2019

#1
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$$\cos(x) - \cos(2x) = 0\\ \cos(x) = \cos(2x)\\ \cos(x) = 2\cos^2(x) -1\\ 2\cos^2(x) - \cos(x) - 1 = 0$$

$$u = \cos(x)\\ 2u^2 - u - 1 = 0\\ (2u+1 )(u -1) = 0\\ u = 1,~-\dfrac 1 2$$

$$\cos(x) = 1 \Rightarrow x = 2k\pi, ~k \in \mathbb{Z}\\ \cos(x) = -\dfrac 1 2 \Rightarrow x = \dfrac{2\pi}{3}+2k\pi,~\dfrac{4\pi}{3}+2k\pi,~k\in \mathbb{Z}$$

$$x = \left \{0,~\dfrac{2\pi}{3},~\dfrac{4\pi}{3}\right\} + 2\pi k,~k \in \mathbb{Z}$$

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Feb 22, 2019