<s>[INST] trigonometry equation - $\sin^3(x)+\sin^3(2x)+\sin^3(3x) = (\sin(x)+\sin(2x)+\sin(3x))^3$ how can i find all possible solutions to this equation?
$\sin^3(x)+\sin^3(2x)+\sin^3(3x) = (\sin(x)+\sin(2x)+\sin(3x))^3$
I've tried writing it all as a sum in the form $\sin^m(x)\cos^n(x) + ... sin(x)$, with just $\sin(x)$ and $\cos(x)$ occuring. For my luck, you can write it in the form $\sin(x)* (...) = 0$ and therefore you can find a few solutions. But inside the brackets there is an expression starting with somthing like $\sin^7(x)$ which i cant handle and dont find any other solutions.
 [/INST] Let $\sin x=a$, $\sin2x=b$,
and $\sin3x=c$
Expand out 
$(a+b+c)^3$
Then subtract $a^3, b^3$ and $c^3$, which yields 
$3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + 3 b^2 c + 3 b c^2=0$.
The long expression can be factored nicely into :
$3(a+b)(b+c)(a+c)$, which equal  to $0$.
The factor $3$ makes no difference, so solve for:
$\sin(x)+\sin(2x)=0$,
$\sin(2x)+\sin(3x)=0$, and
$\sin(x)+\sin(3x)=0$.
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