# Impossible solutions in trigonometric equations

I'm trying to solve $\sin{4v} + \cos{4v} = 0$

I get 4 equations which I can solve for the solutions, including these 2:

$4v_1 = \frac{\pi}{2} + 4v_1 + 2\pi n$

$4v_2 = -\frac{\pi}{2} + 4v_1 + 2\pi n$

These are obviously impossible equations. What does this mean? Why do they show up? It feels like I sometimes have to discard solutions like these because of this fact, but why do they even show up in the first place?

Possibly relevant steps:

$\sin{4v} + \cos{4v} = 0$

$\cos{4v} = -\sin{4v}$

$\cos{4v} = \sin{-4v}$

$\cos{4v} = \cos{\frac{\pi}{2} \pm -4v}$

$4v = \pm (\frac{\pi}{2} \pm 4v)$

• Can you show the intermediate steps? – HDE 226868 Aug 30 '15 at 21:38
• I don't know if it's relevant, but I'll go ahead and add them. @HDE226868 – user262493 Aug 30 '15 at 21:38
• @HDE226868 steps added – user262493 Aug 30 '15 at 21:42
• Why not let $u=4v$ and divide by 4 after solving for $u$? Think about where $\sin(u) = -\cos(u)$ is satisfied on the unit circle... – Chester Aug 30 '15 at 21:47
• Your 4th step, obtained from the 3rd, is incorrect. $\ sin 4v = \cos( \pi /2 -4v)$ )for all $v$. This may be where you introduced the spurious solutions . – DanielWainfleet Aug 31 '15 at 2:15

You could solve it as $\tan 4v=-1$ $$\Rightarrow 4v=-\frac{\pi}{4}+n\pi$$ Etc.
You are correct in discarding these solutions, and from what I know, they are artifacts of a domain change known as extraneous solutions. You've gone from the domain of $\cos {x}$ directly into a first-order polynomial equation, changing your domain from $[-1, 1]$ to $\mathbb{R}$. Your answers are valid for the final equation in $\mathbb{R}$, but stepping back to $[-1, 1]$ you have to shrink your domain and thus discard the extraneous solutions.