Решить уравнение: 3*корень3*sin2x=10sinx-4sin^3*x

$3 \sqrt{3}sin2x = 10sinx - 4sin^3x \\ \\ 6 \sqrt{3} sinxcosx = 2sinx(5 - 2sin^2x) \\ \\ 3 \sqrt{3} sinxcosx - sinx(5 - 2sin^2x) = 0 \\ \\ sinx \cdot (3\sqrt{3} cosx - 5 + 2sin^2x) = 0 \\ \\ sinx = 0 \\ \\ \boxed{x = \pi n, \ n \in Z }$

$3\sqrt{3} cosx - 5 + 2sin^2x = 0 \\ \\ 3\sqrt{3} cosx - 5 + 2 - 2cos^2x = 0 \\ \\ -2cos^2x + 3 \sqrt{3} cosx - 3 = 0 \\ \\ 2cos^2x - 3 \sqrt{3} cosx + 3 = 0 \\ \\$

Пусть $t = cosx, \ t \in [-1; \ 1]$

$2t^2 - 3 \sqrt{3} t + 3 = 0 \\ \\ D = 27 - 4 \cdot 3 \cdot 2 = 27 - 24 = ( \sqrt{3})^2 \\ \\ t_1 = \dfrac{3 \sqrt{3} + \sqrt{3} }{4} = \sqrt{3} - \ \ ne \ \ ud. \\ \\ t_1 = \dfrac{3 \sqrt{3} - \sqrt{3} }{4}= \dfrac{ \sqrt{3} }{2}$

Обратная замена:

$cosx = \dfrac{\sqrt{3} }{2} \\ \\ \boxed{ x = \pm \dfrac{ \pi }{6} + 2 \pi n, \ n \in Z}$