Stucked solving trigonometry equation $\sin \left({{2\,x}\over{3}}+36\right)=\sin 30$

Question

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.

Answer 1

$$\left(\dfrac {2x}{3}+36^\circ\right)= 30^\circ + 360^\circ k \text { (see Note 1)}$$

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

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

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

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).

Answer 2

We can use that

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

with $k\in \mathbb Z$.

Answer 3

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,

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

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

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

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