Question

cos2x-3$\sqrt{3}$cosx+4=0 решить уравнение

Answer 1

$Cos2x-3\sqrt{3}Cosx+4=0\\\\2Cos^{2}x-1-3\sqrt{3}Cosx+4=0\\\\2Cos^{2}x-3\sqrt{3}Cosx+3=0\\\\Cosx=m,-1\leq m\leq 1\\\\2m^{2}-3\sqrt{3}m+3=0\\\\D=(-3\sqrt{3})^{2}-4*2*3=27-24=3=(\sqrt{3})^{2}\\\\m_{1} =\frac{3\sqrt{3}-\sqrt{3}}{4}=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2} \\\\m_{2}=\frac{3\sqrt{3}+\sqrt{3}}{4}=\frac{4\sqrt{3}}{4}=\sqrt{3}>1-neyd\\\\Cosx=\frac{\sqrt{3}}{2} \\\\x=\pm arcCos\frac{\sqrt{3}}{2}+2\pi n,n\in Z\\\\x=\pm\frac{\pi }{6}+2\pi n,n\in Z$