0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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: $$x=-\frac{5 \pi }{6}\lor x=-\frac{2 \pi }{3}\lor x=\frac{\pi }{6}\lor x=\frac{\pi }{3}\lor x=\frac{7 \pi }{6}\lor x=\frac{4 \pi }{3}\lor x=\frac{13 \pi }{6}\lor x=\frac{7 \pi }{3}$$

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

Hint:

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.