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# Find the general solution to cos2x=-cosx

Question
Find the general solution to $$\displaystyle{\cos{{2}}}{x}=-{\cos{{x}}}$$

## Answers (1)

2021-02-12

$$\displaystyle{\cos{{\left({2}{x}\right)}}}={2}{{\cos}^{{2}}{\left({x}\right)}}-{1}$$
$$\displaystyle{2}{{\cos}^{{2}}{\left({x}\right)}}-{1}={\cos{{\left({x}\right)}}}$$
$$\displaystyle{2}{{\cos}^{{2}}{\left({x}\right)}}-{1}-{\cos{{\left({x}\right)}}}={0}$$
$$\displaystyle{2}{{\cos}^{{2}}{\left({x}\right)}}-{\cos{{\left({x}\right)}}}-{1}={0}$$
$$\displaystyle{\left({2}{\cos{{\left({x}\right)}}}+{1}\right)}{\left({\cos{{\left({x}\right)}}}-{1}\right)}={0}$$
$$\displaystyle{2}{\cos{{\left({x}\right)}}}+{1}={0}$$
$$\displaystyle{2}{\cos{{\left({x}\right)}}}=-{1}$$
$$\displaystyle{\cos{{\left({x}\right)}}}=-\frac{{1}}{{2}}$$
$$\displaystyle{x}={{\cos}^{{-{{1}}}}{\left(-\frac{{1}}{{2}}\right)}}$$
$$x = \pi/3 [ 2\pi - \pi/3 = 5\pi/3 ]$$
$$x = \pi/3\ and\ 5 \pi/3$$
cos is positive in the first and fourth quadrant.
$$\displaystyle{\cos{{\left({x}\right)}}}-{1}={0}$$
$$\displaystyle{\cos{{\left({x}\right)}}}={1}$$
$$\displaystyle{x}={{\cos}^{{-{{1}}}}{\left({1}\right)}}$$
$$\displaystyle{x}={0},{2}\pi$$

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