# Solve cos^3x=cosx?

## $C o {s}^{3} x = \cos x$ I get got the $0 , \frac{\pi}{2} , \pi$ but I cant get $\frac{3 \pi}{2}$. If anyone can help out I would really appreciate it.

Feb 10, 2018

See below

#### Explanation:

It's a little tricky, but try setting the equation = 0

${\cos}^{3} x = \cos x$

${\cos}^{3} x - \cos x = 0$

Now factor out $\cos x$

$\cos x \left({\cos}^{2} x - 1\right) = 0$

And now factor the difference of two squares

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

Set each factor equal to 0 and solve using your unit circle

1) $\cos x = 0 , x = \left\{\frac{\pi}{2} , \frac{3 \pi}{2}\right\}$

2) $\cos x + 1 = 0$

$\cos x = - 1 , x = \pi$

3) $\cos x - 1 = 0$
$\cos x = 1 , x = 0$

Feb 11, 2018

See below.

#### Explanation:

${\cos}^{3} \left(x\right) = \cos \left(x\right)$

${\cos}^{3} \left(x\right) - \cos \left(x\right) = 0$

$\cos \left(x\right) \left({\cos}^{2} \left(x\right) - 1\right) = 0$

$\cos \left(x\right) \cdot \left(- 1\right) \left({\cos}^{2} \left(x\right) - 1\right) = 0$

$\cos \left(x\right) \left(- {\cos}^{2} \left(x\right) + 1\right) = 0$

Pythagorean identity:

$\cos \left(x\right) \left({\sin}^{2} \left(x\right)\right) = 0$

$\therefore$

$\cos \left(x\right) = 0 \implies \frac{\pi}{2} , \frac{3 \pi}{2}$

${\sin}^{2} \left(x\right) = 0 \implies 0 , \pi , 2 \pi$