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# How do you solve the following equation cos2x+cosx=0 in the interval [0, 2pi]?

Jul 4, 2016

I found:
$x = \pi$
$x = \frac{\pi}{3} \mathmr{and} \frac{5 \pi}{3}$

#### Explanation:

We can use trigonometric identities to change all into $\cos$ as:
${\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right) + \cos \left(x\right) = 0$
and:
${\cos}^{2} \left(x\right) - 1 + {\cos}^{2} \left(x\right) + \cos \left(x\right) = 0$
$2 {\cos}^{2} \left(x\right) + \cos \left(x\right) - 1 = 0$
We can solve this using the Quadratic Formula as in a second degree equation in $\cos \left(x\right)$;
we can write $\cos \left(x\right) = t$
and we get:
$2 {t}^{2} + t - 1 = 0$
${t}_{1 , 2} = \frac{- 1 \pm \sqrt{1 + 8}}{4} = \frac{- 1 \pm 3}{4}$
${t}_{1} = - 1$
${t}_{2} = \frac{1}{2}$
but: $t = \cos \left(x\right)$ so we have:
$\cos \left(x\right) = - 1$ when $x = \pi$
$\cos \left(x\right) = \frac{1}{2}$ when $x = \frac{\pi}{3} \mathmr{and} \frac{5 \pi}{3}$