Alright, so I know to do this and that this is correct from years of precalc, etc. but I was never taught why. I want to know why and why this does not ALWAYS apply it seems.

When given this problem:

sin x = -1/2

you know that sin is negative in first and 2nd quadrants so the answer is 30 degrees in the terminal angle, so its 30 in the 3rd and 4th quadrants. This make the answer 210 and 330 degrees. I get that and that is what I have answered in past.

On certain equations however (like this one) you need to write:

210 +- 360n, 330 +- 360n

And it seems you do this however many answers you have. WHY do you write +- 360n? What is n, and why would you need to write 360?

Then with something like

4sin^2x = 3

I am more confused, as the answer is clearly 60 degrees, and it is positive, so I would think following these rules the answer would be

60 +- 360n, 120 +- 360n

But because there is a 4 coefficient originally it is

60 +- 360n, 120 +- 360n, 240 +- 360n, 300 +- 360n

Why does this happen?

  • $\begingroup$ Maybe consider to use radians instead of degrees ? $\endgroup$
    – Jean Marie
    Sep 6, 2016 at 1:21
  • $\begingroup$ Even so why would it be +- 2pi*n? $\endgroup$
    – blue
    Sep 6, 2016 at 1:30
  • $\begingroup$ Well, an angle can swing around past 360. So an angle of 380 is the same thing as 20. That's why +/- 360. Swing around the circle n times and you've not 20 + n360. That's why. As for the second question, the 4 has nothing to do with it. The original answer should include, 240 and 300 as well. For some reason you left them out. $\endgroup$
    – fleablood
    Sep 6, 2016 at 2:56

1 Answer 1


In the context of such questions, it is usually understood that $n$ is implicitly an arbitrary natural number.

It signifies that the sine of an angle is equal to the sine of that angle plus an arbitrary number of full rotations (positive or negative). $$\forall n\in\Bbb N:~\sin (\theta) = \sin(\theta \pm 360^\circ n)$$

So to be complete when listing all angles that are the arcsine of a value, you include arbitrary full rotations.

$$\arcsin(-\tfrac 12)\in\{210^\circ \pm 360^\circ n, 330^\circ \pm 360^\circ n\}$$

That is all.


But because there is a 4 coefficient originally it is

No, in that case it was because the sine was squared that there is a base answer in each of the four quadrants.

$$\arcsin \Big(\pm\sqrt{\tfrac 34}\Big) \in \{60^\circ, 120^\circ, 240^\circ, 300^\circ \}\pm\{360^\circ n\}$$

  • $\begingroup$ Ok.. so when it is squared there is always 4? I have seen cases where there are 6 I am fairly sure.. why would that happen? Also I am looking at a squared case where the answer gave only 3 $\endgroup$
    – blue
    Sep 6, 2016 at 1:46
  • $\begingroup$ @skyguy As indicated, there are two square roots for any positive number. The arcsines for the positive square root lie in quadrants 1,2, those for the negative root lie in quadrants 3,4 $\endgroup$ Sep 6, 2016 at 1:51
  • 1
    $\begingroup$ In this case the 4 had nothing to do with it. But if you have something like $sin (4x) = 1/2$ w ith the 4 on the inside then you have 4x = 45+- 360n;135+-360n. Now the 4 matters. x=11.25+- 90n; 33.25+-90n. Note if x = 101.25 then 4x is 405 and sin 405= sin 45 =1/2 so that is an answer. You would have missed it if you didn't have th +-90n. $\endgroup$
    – fleablood
    Sep 6, 2016 at 3:06

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