Solving higher-degree trigonometric equation Is there any method to find value of $\sin (A)$ from $\sin (9A)$ having known value of $\sin (9A) = \sin (30 degree) = 0.5$? 
In $\sin (9A)$, being a nine-degree equation, there will be nine-roots. Can we find all nine roots from $\sin (9A) = 0.5$?
 A: Hint:
from $\sin (9A)=0.5$ we have:
$$
9A= \frac{\pi}{6}+2k\pi \quad \mbox{or}\quad 9A= \frac{5\pi}{6}+2k\pi
$$
with $k \in \mathbb{Z}$. Take different values of $k$ and you can find the values of $A$ simply dividing by $9$.
Note that $\sin (9A)=0.5$ is a trigonometric equation and have infinitely many solutions:
$$
A=\frac{1}{54}\left(\pi + 12k \pi \right)\quad \mbox{or}\quad A=\frac{1}{54}\left(5\pi + 12k \pi \right) \qquad k \in \mathbb{Z}
$$
and there are two groups of $9$ solutions that are repeated with a periodicity of $2\pi$ that we can find from the two expressions using $k\in\{0,1,2,3,4,5,6,7,8\}$
A: Using trigoniometric idendities (and not the way that @EmilioNovati is using):


*

*$$\sin\left(9\text{A}\right)=\sin\left(\text{A}\right)\left(1+2\cos\left(2\text{A}\right)\right)\left(1+2\cos\left(6\text{A}\right)\right)$$

*$$\cos\left(2\text{A}\right)=\cos^2\left(\text{A}\right)-\sin^2\left(\text{A}\right)$$

*$$\cos^2\left(\text{A}\right)=1-\sin^2\left(\text{A}\right)$$

*$$\cos\left(6\text{A}\right)=15\sin^4\left(\text{A}\right)\cos^2\left(\text{A}\right)-15\sin^2\left(\text{A}\right)\cos^4\left(\text{A}\right)+\cos^6\left(\text{A}\right)-\sin^6\left(\text{A}\right)$$

*$$\cos^4\left(\text{A}\right)=\left(1-\sin^2\left(\text{A}\right)\right)^2$$

*$$\cos^6\left(\text{A}\right)=\left(1-\sin^2\left(\text{A}\right)\right)^3$$


So, we get:


*

*$$\sin\left(9\text{A}\right)=\frac{1}{2}\cdot\left(1+2\cos\left(2\text{A}\right)\right)\cdot\left(1+2\cos\left(6\text{A}\right)\right)$$

*$$\cos\left(2\text{A}\right)=1-2\cdot\left(\frac{1}{2}\right)^2=\frac{1}{2}$$

*$$\cos\left(6\text{A}\right)=15\cdot\left(\frac{1}{2}\right)^4\cdot\cos^2\left(\text{A}\right)-15\cdot\left(\frac{1}{2}\right)^2\cdot\cos^4\left(\text{A}\right)+\cos^6\left(\text{A}\right)-\left(\frac{1}{2}\right)^6$$

*$$\cos^2\left(\text{A}\right)=1-\left(\frac{1}{2}\right)^2=\frac{3}{4}$$

*$$\cos^4\left(\text{A}\right)=\left(1-\left(\frac{1}{2}\right)^2\right)^2=\frac{9}{16}$$

*$$\cos^6\left(\text{A}\right)=\left(1-\left(\frac{1}{2}\right)^2\right)^3=\frac{27}{64}$$


So:
$$\sin\left(9\text{A}\right)=\frac{1}{2}\cdot\left(1+2\cdot\frac{1}{2}\right)\cdot\left(1+2\cdot-1\right)=-1$$
