Question

1) решите уравнение
2)найдите корни на промежутке [-2П;(-7П)/2]
$4*sinx*cos^{2} x-2*\sqrt{3} *sin2x+3sinx=0$

Answer 1

$4\cdot Sinx\cdot Cos^{2}x-2\cdot \sqrt{3} \cdot Sin2x+3Sinx =0 \\\\4\cdot Sinx\cdot Cos^{2}x-2\cdot \sqrt{3} \cdot 2Sinx Cosx+3Sinx =0\\\\4SinxCos^{2}x-4\sqrt{3}Sinx Cosx+3Sinx =0\\\\Sinx(4Cos^{2}x-4\sqrt{3}Cosx+3)=0\\\\\\\left[\begin{array}{ccc}Sinx=0\\4Cos^{2}x-4\sqrt{3}x+3=0\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\pi n,n\in Z \\(2Cosx-\sqrt{3})^{2} =0 \end{array}\right$

$\left[\begin{array}{ccc}x=\pi n,n\in Z \\2Cosx-\sqrt{3}=0 \end{array}\right\\\\\\\left[\begin{array}{ccc}x=\pi n,n\in Z \\Cosx=\dfrac{\sqrt{3} }{2} \end{array}\right\\\\\\\left[\begin{array}{ccc}x=\pi n,n\in Z \\x=\pm \dfrac{\pi }{6}+2\pi n,n\in Z \end{array}\right$

Это решение уравнения. Промежуток непонятно записан .