How do I solve this equation $ \sin(x) = \sin(\frac{11}{9}\pi)$?

Question

We are supposed to use this formula for which I can't find any explaination anywhere and our teacher didn't explain anything so if anyone could help me I would appreciate it.

$ x = A + k \times 2\pi$

and

$x = \pi - A + k \times 2\pi$

where $k$ is supposed to be a random integer? and A in this case is $\frac{11}{9}\pi$

Answer 1

The first step is always the same:

DRAW A DIAGRAM!

It doesn't need to be a particularly precise diagram -- we're not going to measure anything on it, just get an overview of how things fit together. Here's a rough graph of $\sin(x)$ with $x=\frac{11}{9}\pi$ marked:

$\frac{11}{9}\pi$ is a bit more than $\pi$, so the value of $\sin \frac{11}{9}$ is somewhere between $-1$ and $0$. We need to find all the $x$ that have that value as their sine.

$x=\frac{11}{9}\pi$ itself is certainly one solution. And we know that the sine repeats itself with a period of $2\pi$, so $x=\frac{11}{9}\pi+2\pi$ is another solution, and $\frac{11}{9}\pi+2\pi+2\pi$ is yet another one, and so forth, and to the other side $\frac{11}{9}\pi-2\pi = -\frac{7}{9}\pi$ is also a solution, and so forth. This gives us $$ x = \frac{11}{9}\pi + 2k\pi, k\in\mathbb Z, $$ the points indicated in green here:

These are all the points where the sine curve passes down through the red horizontal line. We're still missing the ones where the curve crosses back up again, but we can figure out where they are because the sine curve is mirror symmetric about each of it throughs and crests. So the first up crossing after $x=0$ is just as far left of $2\pi$ as our known down crossing is right of $\pi$, and we get for this point $$ x = 2\pi - (\frac{11}{9}\pi - \pi) = \frac{16}{9}\pi $$ This point also has copies at regular $2\pi$ intervals to the right, so this gives the blue solutions $$ x = \frac{16}{9}\pi + 2k\pi, k\in\mathbb Z $$

The complete solution is $$ x = \begin{cases} \frac{11}{9}\pi + 2k\pi & k\in\mathbb Z \\ \frac{16}9 \pi + 2k\pi & k \in\mathbb Z \end{cases} $$

Answer 2

When you have equation $\sin x=a$, the solutions are $x=\arcsin a +2\pi n$ and $x=\pi-\arcsin a + 2\pi n$. This is based on the identity $\sin(\pi-a)=\sin a$. In your case $a=\sin(\frac{11\pi}{9})$ so the solutions are $x=\arcsin(\sin(\frac{11\pi}{9}))+2\pi n=-\frac{2\pi}{9}+2\pi n$ and $x=\pi-\arcsin(\sin(\frac{11\pi}{9}))+2\pi n=\frac{11\pi}{9}+2\pi n$. Term $2\pi n$ is added becase sine is $2 \pi$ periodic thus $\sin(a+2\pi n)=\sin(a)$

Answer 3

Note that you have the arcsine function has a range of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, but sine is negative in both quadrants $3$ and $4$, so only the quadrant $4$ angle is returned. (The opposite is also true: sine is positive in both quadrant $1$ and $2$, but only the quadrant $1$ angle is returned.) Also note that $\sin(\pi-\theta) = \sin(\theta)$.

Hence, we have

$$\sin (x) = y \implies x = \begin{cases}\ \arcsin y+2\pi k \\ \pi-\arcsin y+2\pi k\end{cases}; \quad k\in \mathbb{Z}$$

So, if $$\sin(x) = \sin\left(\frac{11}{9}\pi\right)$$

then there are two cases:

$$x = \begin{cases}\ \frac{11}{9}\pi+2\pi k \\ \pi-\frac{11}{9}\pi+2\pi k\end{cases}$$

with the first being for angles in quadrant $3$ and the second being for angles in quadrant $4$.

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As for why we have $2\pi k$ for $k \in \mathbb{Z}$ ($k$ must be an integer), this is because trig functions are periodic. Note that since $2\pi$ (or $360°$) represents one full revolution, you return back to where you started. For example, $\frac{9\pi}{4}$ radians would mean $2\pi+\frac{\pi}{4}$, so it’s represents the same angle as $\frac{\pi}{4}$. So, any two angles separated by $2\pi$, $4\pi$, $6\pi$, ... radians actually represent the same angle. (In degrees, that would be $360°$, $720°$, $1080°$, etc.) So, we generalize this as all integer multiples of $2\pi$ radians or $360°$.

As a simple example, assume we have $\sin(x) = 1$. Clearly, the answer must be $\pi$. But $\pi+2\pi$ also works. So does $\pi-2\pi$. So does any angle $2\pi k$ radians apart from $\pi$. So we would generalize the answer and give the full solution as $\pi+2\pi k$, as they all represent the same angle, a right angle. In other words, there are infinite solutions.

However, you may occasionally be given a restricted domain such as $0 \leq x \leq 2\pi$. Then, you include only the angles in the four quadrants. (In the example you gave, you omit the $2\pi k$ for both cases, and in the example I gave, the answer simply becomes $\pi$.)