<s>[INST] Solving cubic equations with sine and cosine sums. I was playing with math, and then I tried to rewrite some cubic equation with sine power reduction formula
$$y^3 + my^2 + ny + d = 0.$$
Let
$$y = \sin(x).$$
Then
$$y^2 = \frac{1 - \cos(2x)}{2},$$
$$y^3 = \frac{3\sin(x) - \sin(3x)}{4}.$$
So the equation will be
$$\frac{3\sin(x)-\sin(3x)}{4} + m\frac{1 - \cos(2x)}{2} + n\cdot\sin(x) +d.$$
We can redefine constant multipliers and we will end with
$$a\cdot\sin(x) + b\cdot\cos(2x) + c\cdot\sin(3x) +L = 0.$$
What would be the best approach to solve this without needing to solve polynomial equations whose degree is greater or equal to 3?
 [/INST] This approach has been used to resolve the diminished cubic $x^3 = 3px - 2q$
$x = 2\sqrt p\cos\theta\\
8p\sqrt p\cos^3 \theta = 6p\sqrt p \cos\theta + 2q\\ 
2p\sqrt p(4\cos^3 \theta - 3\cos\theta) = 2q\\
\cos 3\theta = \frac {q}{p\sqrt p}$
We detour into the complex plane if the original cubic does not have 3 real roots.
The substitution $y = x - \frac {m}{3}$ will allow you to turn your initial cubic into a diminished cubic.
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