2 * $\sin(2x+π)$ + $\sqrt{3}$ = 0, Domain : {-π $\le $x $\le $3π}

My working :

u = 2x + π

Changed the domain to -π \le x \le 3π

$\sin(u)$ = -$\sqrt{3} \over 2$

Base Angle is $π \over 3$

So If i have to go -180 degrees:

There are 2 Quadrants that u can be negative. Quadrant 3 and 4.

So I subtracted 2π - π/3 = $5π \over 3$

Then added π + π/3 = $4π \over 3$

Now I asked myself as I do a round of 2π what angles will I hit where sin is negative.

The negative angles are still in Quadrant 3 and 4.

So I added 2π + 5π/3 = $11π \over 3$

then did the same to 4π/3 from which I got 10π/3

Now I did :

2x + π = 5π/3, 4π/3, 11π/3, 10π/3

subtracted the π from all the answers and then divided by 2.

I got :

x = π/3, π/6, 2π/3, 7π/6

Now my answer is actually wrong and I don't understand why. I believe it's due to my understanding of the concept with negative radians in the domain.

Can someone tell me where I'm going wrong?

I looked at this question : How to solve trigonometric equations with a domain involving negative values of $x$?

But i didn't understand the solution especially the part with the variable "m" coming from nowhere and what you're suppose to with that variable.

EDIT: apologies for the format, i don't know what i'm also doing wrong with the mathJax either.

  • $\begingroup$ you must include your text in Dollar signs $\endgroup$ – Dr. Sonnhard Graubner Nov 21 '17 at 16:34
  • $\begingroup$ where should I put the dollar signs? $\endgroup$ – jame_smith Nov 21 '17 at 16:35
  • $\begingroup$ write $$\sin(x)$$ $\endgroup$ – Dr. Sonnhard Graubner Nov 21 '17 at 16:35
  • $\begingroup$ I tried putting a dollar sign near the less than sign, didn't work $\endgroup$ – jame_smith Nov 21 '17 at 16:36
  • 1
    $\begingroup$ Here is a mathjax tutorial for future reference. $\endgroup$ – N.Bach Nov 21 '17 at 16:40

write $$\sin(2x+\pi)=-\frac{\sqrt{3}}{2}$$ and Substitute $$t=2x+\pi$$ then you have to solve $$\sin(t)=-\frac{\sqrt{3}}{2}$$ and this is what i got: $$x=-\frac{5 \pi }{6}\lor x=-\frac{2 \pi }{3}\lor x=\frac{\pi }{6}\lor x=\frac{\pi }{3}\lor x=\frac{7 \pi }{6}\lor x=\frac{4 \pi }{3}\lor x=\frac{13 \pi }{6}\lor x=\frac{7 \pi }{3}$$

  • $\begingroup$ I already did that part, that was one of the first things I did. $\endgroup$ – jame_smith Nov 21 '17 at 16:44
  • $\begingroup$ then you must substitute back and look if your solution is situated in the given interval $\endgroup$ – Dr. Sonnhard Graubner Nov 21 '17 at 16:46
  • $\begingroup$ What do you mean? I did that $\endgroup$ – jame_smith Nov 21 '17 at 16:46
  • $\begingroup$ @jame_smith I believe he meant verifying your solutions. For instance one of your early mistake is to consider $u=\pi+\frac{\pi}3$ a solution to $\sin(u)=-\frac{\sqrt{3}}2$. Edit: if your end solutions are wrong, you must have made a mistake somewhere. If you really have no idea where that mistake is, you can always backtrack your results/reasoning, and check their consistency. $\endgroup$ – N.Bach Nov 21 '17 at 16:50
  • $\begingroup$ but it is, it was like one of the 2 solutions I got right and I did backtrack my results I really don't understand my mistake $\endgroup$ – jame_smith Nov 21 '17 at 16:53


The period of the sine is two pi and supplementary angles have the same sine. So

$$\sin x=s$$ has the solutions

$$2k\pi+\arcsin s, \\(2k+1)\pi-\arcsin s.$$

For your $s$, find all $k$ that make the value fall in the requested range.

  • $\begingroup$ Might be too much to ask but is there a simpler way u can explain it to me cos that just went over my head $\endgroup$ – jame_smith Nov 21 '17 at 16:56
  • $\begingroup$ @jame_smith: plug your $s$ in my equations and try several $k$. $\endgroup$ – Yves Daoust Nov 21 '17 at 17:10

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