Another Trig equation

When you got $\cos 2\theta = \cos (\frac{\pi}{2}-2\theta)$ you would do $2\theta = 2\pi n \pm (\frac{\pi}{2} - 2\theta)$ but in this special case we cannot pick the -ve otherwise we run into nonsense. Instead, we only get the +ve and hence we obtain $\theta = \frac{\pi}{2}n + \frac{\pi}{8}$. Cycling through $n=0,1,2,3$ yields us the four solutions, including the two you are missing.

In other words, we run into nonsense but we do not 'lose' any solutions:
$2\theta = \frac{\pi}{2} - 2\theta \implies \theta = \frac{\pi}{8}$
$2\theta = 2\pi + \frac{\pi}{2} - 2\theta \implies \theta = \frac{5\pi}{8}$
$2\theta = 4\pi + \frac{\pi}{2} - 2\theta \implies \theta = \frac{9\pi}{8}$
$2\theta = 6\pi + \frac{\pi}{2} - 2\theta \implies \theta = \frac{13\pi}{8}$

Thanks again and sorry about these long working trig problems

When you got $\cos 2\theta = \cos (\frac{\pi}{2}-2\theta)$ you would do $2\theta = 2\pi n \pm (\frac{\pi}{2} - 2\theta)$ but in this special case we cannot pick the -ve otherwise we run into nonsense. Instead, we only get the +ve and hence we obtain $\theta = \frac{\pi}{2}n + \frac{\pi}{8}$. Cycling through $n=0,1,2,3$ yields us the four solutions, including the two you are missing.

In other words, we run into nonsense but we do not 'lose' any solutions:
$2\theta = \frac{\pi}{2} - 2\theta \implies \theta = \frac{\pi}{8}$
$2\theta = 2\pi + \frac{\pi}{2} - 2\theta \implies \theta = \frac{5\pi}{8}$
$2\theta = 4\pi + \frac{\pi}{2} - 2\theta \implies \theta = \frac{9\pi}{8}$
$2\theta = 6\pi + \frac{\pi}{2} - 2\theta \implies \theta = \frac{13\pi}{8}$

1 last question, sorry.

How do you think of $n$? Do you think of it as revolutions around a unit circle? or perhaps half revolutions of a unit circle or perhaps something different?