Question

Решите уравнение
4Sin^3x+3sinx+2√3= 2√3cos2x
Отберите корни уравнений на отрезке [2п;7п/2]

Answer 1

$4Sin^{3} x+3Sinx+2\sqrt{3} =2\sqrt{3}Cos2x\\\\4Sin^{3} x+3Sinx+2\sqrt{3} -2\sqrt{3}Cos2x=0\\\\4Sin^{3} x+3Sinx+(2\sqrt{3} -2\sqrt{3}Cos2x)=0\\\\4Sin^{3} x+3Sinx+2\sqrt{3}(1 -Cos2x)=0\\\\4Sin^{3} x+3Sinx+2\sqrt{3}\cdot 2Sin^{2}x =0\\\\4Sin^{3} x+3Sinx+4\sqrt{3}Sin^{2}x =0\\\\Sinx\cdot (4Sin^{2}x +4\sqrt{3}Sinx+3)=0\\\\1)Sinx=0\\\\x=\pi n,n\in Z\\\\n=2 \ \Rightarrow \ x_{1} =2\pi \\\\n=3\\\\x_{2}=3\pi\\\\2)4Sin^{2} x+4\sqrt{3} Sinx+3=0\\\\(2Sinx+\sqrt{3})^{2}=0\\\\2Sinx+\sqrt{3} =0$

$Sinx=-\dfrac{\sqrt{3} }{2}\\\\\left[\begin{array}{ccc}x=-\dfrac{\pi }{3}+2\pi n,n\in Z \\x=-\dfrac{2\pi }{3}+2\pi n,n\in z\end{array}\right\\\\1)x=-\dfrac{\pi }{3}+2\pi n,n\in Z \\\\2)x=-\frac{2\pi }{3} +2\pi n,n\in Z\\\\n=2 \ \Rightarrow \ x_{3}=\frac{10\pi }{3} \\\\Otvet:\boxed{2\pi \ ; \ 3\pi \ ; \frac{10\pi }{3} }$