<s>[INST] Trigonometric equation $\sin v = -1/\sqrt{2}$ I'm trying to solve the following: $$\sin(v) = -\frac{1}{\sqrt{2}}$$
My attempt: 
$$-\sin(v) = \frac{1}{\sqrt{2}}$$
$$\sin(-v) = \frac{1}{\sqrt{2}}$$
$$v_1 = -\frac{\pi}{4} - 2\pi n $$
$$v_2 = -\pi + \frac{\pi}{4} - 2\pi n = \frac{-3\pi}{4} - 2\pi n$$
$v_1 $ is correct, but $v_2 $ should apparently be $ \frac{5\pi}{4} + 2\pi n$
why?
 [/INST] Observe that $$-\frac{3\pi}{4} + 2\pi = \frac{5\pi}{4}$$  If you substitute $-m - 1$ for $n$ in the expression $$-\frac{3\pi}{4} - 2\pi n, n \in \mathbb{Z}$$ you obtain $$-\frac{3\pi}{4} - 2\pi(-m - 1) = -\frac{3\pi}{4} + 2\pi m + 2\pi = \frac{5\pi}{4} + 2\pi m$$
where $m = -n - 1 \in \mathbb{Z}$.
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