solve $\sin(3x)\cos(6x)-\cos(3x)\sin(6x)=-.9$ 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.  
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.  $$\frac13(\arcsin(.9)), \frac{\pi}{3}-\frac13(\arcsin(.9)), \frac{2\pi}{3}+\frac13(\arcsin(.9)), \pi - \frac13(\arcsin(.9)),$$ $$\frac{4\pi}{3}+\frac13(\arcsin(.9)), \frac{5\pi}{3}-\frac13(\arcsin(.9))$$
 A: 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 $3 \cdot \frac{2\pi}3 = 2\pi)$
A: 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

*

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


*$~\displaystyle \frac{\theta_2}{3} + \frac{2k\pi}{3} ~: ~k \in \{0,1,2\}.$
A: 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 $\theta = \pi - \varphi$.

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 $[0, 2\pi)$.
