# 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?

This approach has been used to resolve the diminished cubic $$x^3 = 3px - 2q$$
$$x = 2\sqrt p\cos\theta\\ 8p\sqrt p\cos^3 \theta = 6p\sqrt p \cos\theta + 2q\\ 2p\sqrt p(4\cos^3 \theta - 3\cos\theta) = 2q\\ \cos 3\theta = \frac {q}{p\sqrt p}$$
The substitution $$y = x - \frac {m}{3}$$ will allow you to turn your initial cubic into a diminished cubic.