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**Prove It** I take it the angle is measured in degrees...

$\displaystyle 8\cos^3{10} - 6\cos{10} = 2(4\cos^3{10} - 3\cos{10})$

$\displaystyle = 2\cos{(3\cdot 10)}$

$\displaystyle = 2\cos{30}$

$\displaystyle = \frac{2\sqrt{3}}{2}$

$\displaystyle = \sqrt{3}$.

$\displaystyle 8\cos^3{x} - 6\cos{x} = -1$

$\displaystyle 2\cos{3x} = -1$

$\displaystyle \cos{3x} = -\frac{1}{2}$

Since cosine is negative in the second and third quadrants...

$\displaystyle 3x = \{180 - 60, 180 + 60\} + 360n, n \in \mathbf{Z}$

$\displaystyle 3x = \{120, 240\} + 360n$

$\displaystyle x = \{40, 80\} + 120n$

In the region $\displaystyle 0 < x < 180, x = \{40, 80, 160\}$.