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I am practice about find the x as solution of $$\sin \left({{2\,x}\over{3}}+36^\circ\right)={{1}\over{2}}$$ Between $0^\circ$ and $360^\circ$ Here is my work so far, $$\sin \left({{2\,x}\over{3}}+36^\circ\right)=\sin 30^\circ$$ I try using arcsin $${{2\,x}\over{3}}+36^\circ=30^\circ$$ $${2x\over3}=-6^\circ$$ $$x=-9^\circ$$ I am stucked here. I do not know negative angle appear here, what should i do next ? Thank you for your help and suggestion.

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  • $\begingroup$ if you have $\sin \left({{2\,x}\over{3}}+36^\circ\right)=\sin 30^\circ$ why you don't use the $arcsin$ function in both sides of the equation and see what happen in $0<x<360$? $\endgroup$ – sango Sep 29 '19 at 12:17
  • $\begingroup$ Thank you for your suggestion iwill try now. $\endgroup$ – bambang nugroho Sep 29 '19 at 12:20
  • $\begingroup$ $\sin 30º = \frac{1}{2}$ which is positive. Using CAST (which is a sign diagram for the four quadrants), $\sin x$ is positive both in the first quadrant and the third quadrant. So you actually need to solve $\frac{2(x+180º)}{3} + 36º = 30º$. In addition, since $\sin(x + 360º) = \sin x$, you need to solve $\frac{2(x+360º)}{3} + 36º = 30º$ as well. $\endgroup$ – Toby Mak Sep 29 '19 at 12:43
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$$\left(\dfrac {2x}{3}+36^\circ\right)= 30^\circ + 360^\circ k \text { (see Note 1)}$$

$$\left(\dfrac {2x}{3}\right)= -6^\circ + 360^\circ k$$

$$2x = -18^\circ + 1080^\circ k $$

$$x = -9^\circ + 540^\circ k$$

However, since $540^\circ k = 180^\circ k$

$$x = -9^\circ + 180^\circ k$$

Thus your solutions are (using $k=1$ and $k=2$ $\text {(see Note 2)}$

$$x = 171^\circ, x = 351^\circ$$

Note 1: Whenever you're solving for a range, add $360^\circ k$ if working in degrees or $2 \pi k$ if working in radians.

Side note: don't mix up the two angular measures; that is, don't add $360^\circ k$ if working in radians (and don't add $2\pi k$ if working in degrees).

Note 2: I explicitly left out $k=0$ because that solution would give us $-9^\circ$ - since we're looking for a positive solution, the only solutions that give positive angles are $k=1$ and $k=2$, which are $171^\circ$ and $351^\circ$.

Side note #2: If we measure the angles in the clockwise direction, $-9^\circ$ and $-189^\circ$ are the correct solutions (in fact, they are coterminal with the positive counterclockwise angles).

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    $\begingroup$ Thank you so much for your explanation, suggestion and information. I get more knowledge from your explanation. Working like this problem always add $360^\circ×k$. I always remember that. Thank you so much. $\endgroup$ – bambang nugroho Sep 29 '19 at 17:11
  • $\begingroup$ My pleasure! I've expanded my explanations above. $\endgroup$ – bjcolby15 Sep 29 '19 at 22:32
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In the unit circle definition of trigonometry, If the point $(x,y)$ corresponds to the angle $\theta$, written $P(\theta)=(x,y)$, then $\sin \theta = y$.

There are two points on the unit circle that have the same $y$-coordinate, $P(\theta)$ and $P(\pi - \theta)$.

Also, in general, for any point, $\theta$, $P(\theta) = P(\theta + 2n\pi)$, for all integers, $n$.

So if $\sin A = \sin B$, then

$$B = A + 2n\pi \qquad \text{or} \qquad B = (\pi - A) + 2n \pi $$

In your case,

$$\sin \left({{2\,x}\over{3}}+36^\circ\right) =\sin(30^\circ + 360^\circ n) \quad \text{or} \quad = \sin(150^\circ + 360^\circ n)$$

$$\dfrac 23 x+36^\circ =30^\circ + 360^\circ n \quad \text{or} \quad = 150^\circ + 360^\circ n$$

$$\dfrac 23 x =-6^\circ + 360^\circ n \quad \text{or} \quad = 114^\circ + 360^\circ n$$

$$x =-9^\circ + 540^\circ n \quad \text{or} \quad = 171^\circ + 540^\circ n$$

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  • $\begingroup$ Unit circle is all what we need for trigonometry! $\endgroup$ – user Sep 29 '19 at 12:42
  • $\begingroup$ Thank you for your suggestion, i will try it now. $\endgroup$ – bambang nugroho Sep 29 '19 at 12:44
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We can use that

$$\sin A=\sin B\iff A=B+2\pi k \;\lor A =\pi-B+2\pi k $$

with $k\in \mathbb Z$.

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  • $\begingroup$ I do not know Z here. And what is the k for, can you explain futher more ? Thank you so much. $\endgroup$ – bambang nugroho Sep 29 '19 at 12:33
  • $\begingroup$ Since trigonometric are periodic, the term $2k\pi$ represents in radians all the integer multiple of 360 degree angle. If you are interested to the general solution you need to take into account that otherwise you can choose the correct value of k to obtain the solution for the desidered interval. $\endgroup$ – user Sep 29 '19 at 12:38

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