### Instruction: Find all $a,$ $0^\circ < a < 360^\circ,$ such that $\cos a,$ $\cos 2a,$ and $\cos 3a$ form an arithmetic sequence, in that order.  Enter the solutions, separated by commas, in degrees.### Response: To form an arithmetic sequence, the difference between any two consecutive terms must be constant.
So I need to find $a$ such that $\cos 2a - \cos a = \cos 3a - \cos 2a.$
I can use some trigonometric identities to simplify this equation.
I know that $\cos 2a = 2 \cos^2 a - 1$ and $\cos 3a = 4 \cos^3 a - 3 \cos a.$
So the equation becomes $2 \cos^2 a - 1 - \cos a = 4 \cos^3 a - 3 \cos a - (2 \cos^2 a - 1),$ or equivalently, $4 \cos^3 a - 4 \cos^2 a - 2 \cos a + 2 = 0.$
This looks like a cubic equation, but I can factor it by grouping.
I notice that $4 \cos^3 a - 4 \cos^2 a = 4 \cos^2 a (\cos a - 1)$ and $-2 \cos a + 2 = -2 (\cos a - 1).$
So the equation is equivalent to $(4 \cos^2 a - 2)(\cos a - 1) = 0.$
This means that either $\cos a - 1 = 0$ or $4 \cos^2 a - 2 = 0.$
Solving for $\cos a,$ I get $\cos a = 1$ or $\cos a = \pm \frac{\sqrt{2}}{2}.$
Now I need to find the values of $a$ that correspond to these cosines.
I know that $\cos a = 1$ only when $a = 0^\circ$ or $a = 360^\circ,$ but these are not in the interval $(0^\circ, 360^\circ),$ so I can ignore them.
I also know that $\cos a = \frac{\sqrt{2}}{2}$ when $a = 45^\circ$ or $a = 315^\circ,$ and that $\cos a = -\frac{\sqrt{2}}{2}$ when $a = 135^\circ$ or $a = 225^\circ.$