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I made a graph of an equation and also solved the equation algebraically. Even though I can find one answer, I am having a hard time finding all the rest. All of the answers have to fall in the interval $[0, 2\pi)$. Here is my math and graph

$$\sin(3x)\cos(6x)-\cos(3x)\sin(6x)=-.9$$ I used the difference of sin trig identity. $$\sin(3x-6x)=-.9$$ $$\sin(-3x)=-.9$$ I used the even and odd trig identity $$-\sin(3x)=-.9$$ I did graph this equation because I was lost when looking at the book answers. I labeled the points that may have some relevance. The period $\frac{2\pi}{3}$ seemed important, there are three points in the graph that show it equal to $.9$. I list all the answers at the end of the post and the points labeled on the graph show up in the answer. But, that is as close to the answer I can come. y=-sin(3x)+.9

The book has the answer in $\arcsin$. I solved the equation for that. $$-\arcsin(-.9)=3x$$ I used the even and odd identity for $\sin$. $$\arcsin(.9)=3x$$ Then I divided both sides by $3$ and this gave me the first answer the book was looking for $$\frac13(\arcsin(.9))$$ I realize that since the answer is looking for $3x$ that I have to go around the circle three times. The book says, "Whenever we solve a problem in the form of $\sin(nx) =c$ we must go around the circle $n$ times". This is why I started to graph. I am not sure if the graph has any purpose but I was able to find some points of importance to me. Here are all the answers the book list for the solution. I was wondering if someone could help me in understanding the answer. $$\frac13(\arcsin(.9)), \frac{\pi}{3}-\frac13(\arcsin(.9)), \frac{2\pi}{3}+\frac13(\arcsin(.9)), \pi - \frac13(\arcsin(.9)),$$ $$\frac{4\pi}{3}+\frac13(\arcsin(.9)), \frac{5\pi}{3}-\frac13(\arcsin(.9))$$

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    $\begingroup$ Still reading and thinking, but regarding the graph it would appear you graphed the equation $-\sin(3x) + 0.9,$ so the solutions you're looking for here would be the zeroes on that graph, not the points where it equals $0.9.$ $\endgroup$ Commented Apr 22, 2022 at 2:46
  • $\begingroup$ That’s right. I forgot that I set the equation equal to zero when I put the equation into standard form. May be that was one of my disconnects? Thank you $\endgroup$
    – Benp404
    Commented Apr 22, 2022 at 16:27

3 Answers 3

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We can start looking for the other solutions by considering a substitution using the identity $\sin(\pi - x) = \sin(x).$ This gives us that $\sin(\pi - 3x) = 0.9,$ so $$\pi - 3x = \arcsin(0.9) \Rightarrow 3x = \pi - \arcsin(0.9) \Rightarrow x = \frac\pi3 - \frac13 \arcsin(0.9)$$

gives us the answer where $3x$ is in the second quadrant for the first period. Note that the reason we don't get that without considering this substitution is because any answers for $3x$ in the second or third quadrants are outside of the range of arcsine $(-\frac\pi2, \frac\pi2)$ but because their differences with $\pi$ are inside the range, this substitution shows them to us.

Now because a line can only intersect a circle in at most two places (think about the unit circle) this is the only other solution within one period of $\sin(3x),$ so to get the others we can add the period to either of our solutions. As you pointed out, the period is $\frac{2\pi}3$ because $\sin(3(x + \frac{2\pi}3)) = \sin(3x + 2\pi) = \sin(3x).$ So, in order to get the other solutions, add $\frac{2\pi}3$ to either solution either once or twice. (three times would result in a coterminal angle because $3 \cdot \frac{2\pi}3 = 2\pi)$

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  • $\begingroup$ Stephen - Thanks for a detailed answer! $\endgroup$
    – Sundar
    Commented Apr 22, 2022 at 3:11
  • $\begingroup$ The range of arcsine is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. $\endgroup$ Commented Apr 22, 2022 at 8:54
  • $\begingroup$ Thanks. I was wondering how to get the other angle. One mystery solves. I will have to find more meaning in the range and the other solution where the line intersects the unit circle. Thank you. $\endgroup$
    – Benp404
    Commented Apr 22, 2022 at 16:24
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Alternative perspective:
An intuitive approach.

You want to find all values $x$ such that $\sin(3x) = 0.9$, with $0 \leq x < 2\pi.$

Suppose that $\theta = 3x$.

Then, a starting point would be to identify all values $\theta$ such that $0 \leq \theta < 2\pi$ and $\sin(\theta) = 0.9.$

Rather than examining the graph of $y = \sin(x)$, I find it helpful to visualize the unit circle, and consider what happens as $\theta$ rotates from $0$ through $2\pi$ around this unit circle.

Clearly, there will be exactly one value of $\theta$ in the first quadrant such that $\sin(\theta) = 0.9.$

Refer to this angle as $\theta_1$.

Further, by symmetry, you will have exactly one value of $\theta$ in the 2nd quadrant such that $\sin(\theta) = 0.9$.

Refer to this angle as $\theta_2$.

From the visualization of the sine function, against the backdrop of the unit circle, you know that for any angle $\theta$ such that $\pi \leq \theta < 2\pi$, you will have that $\sin(\theta) < 0.$

Therefore, $\theta_1$ and $\theta_2$ are the only $2$ angles, within a modulus of $2\pi$, such that $\sin(\theta) = 0.9.$

So, before preceding further, some method is needed to be able to explicitly refer to $\theta_1$ and $\theta_2$.

The Arcsine function has as its range, $-\pi/2 \leq \theta \leq \pi/2,$ within a modulus of $2\pi.$ So, the Arcsine function ranges from quadrant 4 through quadrant 1.

Therefore, $\theta_1 = \text{Arcsine}(0.9).$

Then, either by a symmetrical visualization, or by consideration of the formula :
$\sin(\pi - \theta_1) = \sin(\pi)\cos(\theta_1) - \sin(\theta_1)\cos(\pi) = \sin(\theta_1)$

you can intuit that

$\theta_2 = \pi - \theta_1 = \pi - \text{Arcsine}(0.9).$


So, now you have $2$ pertinent angles, within a modulus of $2\pi$

  • $\theta_1 = \text{Arcsine}(0.9).$
  • $\theta_2 = \pi - \text{Arcsine}(0.9).$

Now, you have to find all angles $x$ so that either:

  • $(3x) \equiv \theta_1 \pmod{2\pi}.$
  • $(3x) \equiv \theta_2 \pmod{2\pi}.$

I realize that since the answer is looking for 3x that I have to go around the circle three times. The book says, "Whenever we solve a problem in the form of sin(nx)=c we must go around the circle n times".

This is true but confusing. It is not going to do you much good to try to memorize such a rule. Instead, you need to dissect the rule, and stretch your intuition to understand the thinking behind such a rule.

A better approach is as follows:

  • First, you want to identify all values of $x$ such that $3x \equiv \theta_1 \pmod{2\pi}$.

  • Then, you want to do the same thing for $\theta_2.$

Reason it out.

Suppose that you have some angle $\alpha$ such that

  • $0 \leq \alpha < 2\pi$
  • You want $nx \equiv \alpha \pmod{2\pi} ~: ~n \in \Bbb{Z^+}.$

This means that you want $(nx - \alpha)$ to be a multiple of $2\pi$. Therefore, you want there to exist some integer $k$ so that

$\displaystyle nx = (\alpha + 2k\pi) \implies x = \frac{\alpha + 2k\pi}{n}.$

So, forget about elegance.
Forget about memorizing and blindly following a formula.

Instead, just experiment.

If $k = 0$, then $~\displaystyle x = \frac{\alpha}{n}.$

If $k = 1$, then $~\displaystyle x = \frac{\alpha}{n} + \frac{2\pi}{n}.$

If $k = 2$, then $~\displaystyle x = \frac{\alpha}{n} + \frac{4\pi}{n}.$

If $k = 3$, then $~\displaystyle x = \frac{\alpha}{n} + \frac{6\pi}{n}.$

$\cdots$

If $k = (n-2)$, then $~\displaystyle x = \frac{\alpha}{n} + \frac{2(n-2)\pi}{n}.$

If $k = (n-1)$, then $~\displaystyle x = \frac{\alpha}{n} + \frac{2(n-1)\pi}{n}.$

If $k = (n)$, then $~\displaystyle x = \frac{\alpha}{n} + \frac{2(n)\pi}{n}.$

If $k = (n+1)$, then $~\displaystyle x = \frac{\alpha}{n} + \frac{2(n+1)\pi}{n}.$

At this point, stretching your intuition, you realize that you have let $k$ range through all of the elements in $\{0,1,2,\cdots,(n-2),(n-1),n,(n+1)\}$, and you could have kept going.

This is the AHA moment.
Your intuition tells you that having $k = 0$ and having $k = n$ both repeated the same angle. This is because

$\displaystyle \frac{\alpha}{n}~$ is equivalent to $~\displaystyle \frac{\alpha}{n} + \frac{2n\pi}{n},~$ within a modulus of $(2\pi).$

Similarly, your intuition tells you that $\displaystyle \frac{\alpha}{n} + \frac{2\pi}{n}~$ is equivalent to $~\displaystyle \frac{\alpha}{n} + \frac{2(n+1)\pi}{n},~$ within a modulus of $(2\pi).$

So, you realize that as $k$ went from $0$ through $(n-1)$, you constructed $n$ different angles. Then, as $k$ took on the values of $n$ and $(n+1)$, you saw that the angles were starting to repeat.

So, this is the thinking behind the rather obtusely worded: "Whenever we solve a problem in the form of sin(nx)=c we must go around the circle n times."


In this problem, you have $n=3$ and two separate angles to consider:

  • $\theta_1 = \text{Arcsine}(0.9).$
  • $\theta_2 = \pi - \text{Arcsine}(0.9).$

Now, you have to find all angles $x$ so that either:

  • $(3x) \equiv \theta_1 = \text{Arcsine}(0.9) \pmod{2\pi}.$
  • $(3x) \equiv \theta_2 = \pi - \text{Arcsine}(0.9)\pmod{2\pi}.$

Therefore, the pertinent values for $x$ must be

  • $~\displaystyle \frac{\theta_1}{3} + \frac{2k\pi}{3} ~: ~k \in \{0,1,2\}.$

  • $~\displaystyle \frac{\theta_2}{3} + \frac{2k\pi}{3} ~: ~k \in \{0,1,2\}.$

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  • $\begingroup$ I have been trying to follow the logic and apply it. It takes me so much practice. There is this idea of a series that I am struggling to learn and this post brings about thoughts on how series are used in math. My thoughts on series are not ver developed so this post helps me and gives me more confidence. Madhava is the first mathematician I found in my book that approximates sin through a series. Not sure how it is done yet, but will find a way somehow. Thanks $\endgroup$
    – Benp404
    Commented Apr 22, 2022 at 19:35
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We wish to solve the equation $\sin(3x)\cos(6x) - \sin(6x)\cos(3x) = -0.9$ in the interval $[0, 2\pi)$. \begin{align*} \sin(3x)\cos(6x) - \sin(6x)\cos(3x) & = -0.9\\ \sin(3x - 6x) & = -0.9 && \text{since $\sin\alpha\cos\beta - \cos\alpha\sin\beta = \sin(\alpha - \beta)$}\\ \sin(-3x) & = -0.9 && \text{subtract}\\ -\sin(3x) & = -0.9 && \text{since $\sin(-\theta) = -\sin\theta$}\\ \sin(3x) & = 0.9 && \text{multiply both sides of the equation by $-1$} \end{align*} A particular solution of this equation is \begin{align*} 3x & = \arcsin(0.9)\\ \end{align*} The range of the arcsine function is $[-\pi/2, \pi/2]$. Since $0 < 0.9 < 1$, $0 < \arcsin(0.9) < \pi/2$.

Before we continue, let's consider when $\sin\theta = \sin\varphi$. Clearly, the equation is true if $\theta = \varphi$. Since $\sin\theta$ is the $y$-coordinate of the point where the terminal side of an angle in standard position (vertex at the origin, initial side on the positive $x$-axis) intersects the unit circle, by symmetry, another solution is $\theta = \pi - \varphi$.

symmetry_diagram_for_sine_and_cosine

Moreover, any angle coterminal with one of these angles satisfies the equation $\sin\theta = \sin\varphi$. Hence, $\sin\theta = \varphi$ if $$\theta = \varphi + 2k\pi, k \in \mathbb{Z}$$ or $$\theta = \pi - \varphi + 2m\pi, m \in \mathbb{Z}$$

For the equation above, that means the general solution is \begin{align*} 3x & = \arcsin(0.9) + 2k\pi, k \in \mathbb{Z} & \pi - 3x & = \arcsin(0.9) + 2m\pi, m \in \mathbb{Z}\\ x & = \frac{\arcsin(0.9)}{3} + \frac{2k\pi}{3}, k \in \mathbb{Z} & 3x - \pi & = -\arcsin(0.9) - 2m\pi, m \in \mathbb{Z}\\ & & 3x & = \pi - \arcsin(0.9) - 2m\pi, m \in \mathbb{Z}\\ & & x & = \frac{\pi}{3} - \frac{\arcsin(0.9)}{3} - \frac{2m\pi}{3}, m \in \mathbb{Z}\\ & & x & = \frac{\pi}{3} - \frac{\arcsin(0.9)}{3} + \frac{2n\pi}{3}, n \in \mathbb{Z} \end{align*} where we let $n = -m$ in the preceding step.

We stated above that $0 < \arcsin(0.9) < \dfrac{\pi}{2}$. Hence, $0 < \dfrac{\arcsin(0.9)}{3} < \dfrac{\pi}{6}$. Therefore, $$\frac{\pi}{3} > \frac{\pi}{3} - \frac{\arcsin(0.9)}{3} > \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$

With that in mind, we select values of $k$ and $n$ such that \begin{align*} x & = \frac{\arcsin(0.9)}{3} + \frac{2k\pi}{3}, k \in \mathbb{Z} & x & = \frac{\pi}{3} - \frac{\arcsin(0.9)}{3} + \frac{2n\pi}{3}, n \in \mathbb{Z} \end{align*} are in the interval $[0, 2\pi)$. Those values are $k = 0, 1, 2$ and $n = 0, 1, 2$, which correspond to the solutions \begin{align*} x & = \begin{cases} \dfrac{\arcsin(0.9)}{3},\\[2 mm] \dfrac{\arcsin(0.9)}{3} + \dfrac{2\pi}{3},\\[2 mm] \dfrac{\arcsin(0.9)}{3} + \dfrac{4\pi}{3} \end{cases} & x & = \begin{cases} \dfrac{\pi}{3} - \dfrac{\arcsin(0.9)}{3},\\[2 mm] \dfrac{\pi}{3} - \dfrac{\arcsin(0.9)}{3} + \dfrac{2\pi}{3},\\[2 mm] \dfrac{\pi}{3} - \dfrac{\arcsin(0.9)}{3} + \dfrac{4\pi}{3} \end{cases}\\ & & x & = \begin{cases} \dfrac{\pi}{3} - \dfrac{\arcsin(0.9)}{3},\\[2 mm] \pi - \dfrac{\arcsin(0.9)}{3},\\[2 mm] \dfrac{5\pi}{3} - \dfrac{\arcsin(0.9)}{3} \end{cases} \end{align*} We get six solutions since there are two solutions within each period and $\sin(3x)$ has three periods within the interval $[0, 2\pi)$.

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  • $\begingroup$ Thanks. I am trying to get caught up with you tonight. $\endgroup$
    – Benp404
    Commented Apr 22, 2022 at 16:29
  • $\begingroup$ It seems so strange to think there are multiple periods in one revolution or interval $(0, 2\pi]$ $\endgroup$
    – Benp404
    Commented Apr 22, 2022 at 19:30

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