# How do you solve 2cos^2x+cosx=0 and find all exact general solutions?

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25
Dec 16, 2016

The solutions are $S = \left\{\frac{\pi}{2} + 2 \pi n , \frac{3}{2} \pi + 2 \pi n , \frac{2}{3} \pi + 2 \pi n , \frac{4}{3} \pi + 2 \pi n\right\}$

#### Explanation:

The equation is

$2 {\cos}^{2} + \cos x = 0$

$\cos x \left(2 \cos x + 1\right) = 0$

$\cos x = 0$ and $\cos x = - \frac{1}{2}$

$\cos x = 0$ gives $x = \frac{\pi}{2} + 2 \pi n$ and $x = \frac{3}{2} \pi + 2 \pi n$

$\cos x = - \frac{1}{2}$ gives $x = \frac{2}{3} \pi + 2 \pi n$ and $x = \frac{4}{3} \pi + 2 \pi n$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Dec 16, 2016

General Solution for $2 {\cos}^{2} x + \cos x = 0$ is

$x = \frac{\left(2 n + 1\right) \pi}{2}$ or $x = 2 n \pi \pm \frac{2 \pi}{3}$, where $n$ is an integer.

#### Explanation:

$2 {\cos}^{2} x + \cos x = 0$

$\Leftrightarrow \cos x \left(2 \cos x + 1\right) = 0$

i.e. either $\cos x = 0$ or $2 \cos x + 1 = 0$ i.e. $\cos x = - \frac{1}{2}$

General solution for $\cos x = 0$ is $x = \frac{\left(2 n + 1\right) \pi}{2}$, where $n$ is an integer

and general solution for $\cos x = - \frac{1}{2} = \cos \left(\pm \frac{2 \pi}{3}\right)$ is $x = 2 n \pi \pm \frac{2 \pi}{3}$, where $n$ is an integer

Hence, General Solution for $2 {\cos}^{2} x + \cos x = 0$ is

$x = \frac{\left(2 n + 1\right) \pi}{2}$ or $x = 2 n \pi \pm \frac{2 \pi}{3}$, where $n$ is an integer.

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