<s>[INST] Show that sin(3x) = sin(x)(2cos(2x) + 1) I'm currently studying Trigonometric Functions and came to exercise where I am to find all possible values for x in the range of x[0, PI] where.
The equation is as follows:
sin(3x) = 0.39
I found two values possible for x but the solution in my exercise book shows 2 extra solutions.
Wolfram Alpha tells me that an alternate form of sin(3x) is sin(x)(2cos(2x)+1)
What I want to know is: how do I transform sin(3x) to be the alternate form, and how do I find all the possible solutions for x?
The measurements are in radians.
 [/INST] $$\begin{align}\sin 3x &= \sin(x+2x)\\
&=\sin x\cos 2x + \sin 2x\cos x\\
&=\sin x\cos 2x +2\sin x\cos^2 x \\
&=\sin x(\cos 2x +2\cos^2x)\end{align}$$
But then remember $2\cos^2 x-1=\cos 2x$, so $2\cos^2 x=\cos 2x+1$
In general, when $n$ is an integer:
$$\frac{\sin nx}{\sin x}$$
can be written as a polynomial of degree $n-1$ in $\cos x$. This polynomial is called a Chebyshev polynomial of the second type. It turns out, you can write it instead in terms of a linear combination of  $\cos(n-1)x,\cos(n-3)x,\dots$.
So $$\begin{align}\frac{\sin 2x}{\sin x} &= 2\cos x\\
\frac{\sin 3x}{\sin x} &= 2\cos 2x + 1\\
\frac{\sin 4x}{\sin x} &= 8\cos^3 x -4\cos x=2\cos 3x +2\cos x
\end{align}$$
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