<s>[INST] 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.  $$\frac13(\arcsin(.9)), \frac{\pi}{3}-\frac13(\arcsin(.9)), \frac{2\pi}{3}+\frac13(\arcsin(.9)), \pi - \frac13(\arcsin(.9)),$$ $$\frac{4\pi}{3}+\frac13(\arcsin(.9)), \frac{5\pi}{3}-\frac13(\arcsin(.9))$$
 [/INST] 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 $[0, 2\pi)$.
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