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. $$\frac13(\arcsin(.9)), \frac{\pi}{3}-\frac13(\arcsin(.9)), \frac{2\pi}{3}+\frac13(\arcsin(.9)), \pi – \frac13(\arcsin(.9)),$$ $$\frac{4\pi}{3}+\frac13(\arcsin(.9)), \frac{5\pi}{3}-\frac13(\arcsin(.9))$$
Best Answer
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 $3 \cdot \frac{2\pi}3 = 2\pi)$