Question

Решите уравнение

$\tt\displaystyle\frac{\sin{2x}-2\sin{x}-\sqrt{3}\cos{x}+\sqrt{3}}{tg{x}-\sqrt{3}}=0$

Answer 1

$\frac{Sin2x-2Sinx-\sqrt{3}Cosx+\sqrt{3}}{tgx-\sqrt{3}}=0\\\\\frac{(2Sinx Cosx-2Sinx)-\sqrt{3}(Cosx-1)}{tgx-\sqrt{3}}=0\\\\\frac{2Sinx(Cosx-1)-\sqrt{3}(Cosx-1)}{tgx-\sqrt{3}}=0\\\\\frac{(Cosx-1)(2Sinx-\sqrt{3})}{tgx-\sqrt{3}} =0\\\\\left \{ {{(Cosx-1)(2Sinx-\sqrt{3})=0 } \atop {tgx-\sqrt{3}\neq 0}} \right.$

$1)Cosx-1=0\\\\Cosx=1\\\\x=2\pi n,n\in Z\\\\2)2Sinx-\sqrt{3}=0\\\\2Sinx=\sqrt{3}\\\\Sinx=\frac{\sqrt{3}}{2} \\\\\left[\begin{array}{ccc}x=\frac{\pi }{3}+2\pi n,n\in Z \\x=\frac{2\pi }{3}+2\pi n,n\in Z \end{array}\right\\\\x=\frac{\pi }{3}+2\pi n-neyd,tak...kak tg(\frac{\pi }{3}+2\pi n)=tg\frac{\pi }{3} =\sqrt{3}\\\\Otvet:\boxed{2\pi n;\frac{2\pi }{3}+2\pi n,n\in Z}$

Answer 2

Объяснение:

$\frac{sin(2x)-2sinx-\sqrt{3}*cosx+\sqrt{3} }{tgx-\sqrt{3} }=0$

ОДЗ: tgx-√3≠0       tgx≠√3       x≠π/3+πn.

$sin(2x)-2*sinx-\sqrt{3}*cosx+\sqrt{3} =0\\2*sinx*cosx-2*sinx-\sqrt{3}*cosx+\sqrt{3}=0\\ 2*sinx*(cosx-1)-\sqrt{3}*(cosx-1)=0\\ (cosx-1)*(2*sinx-\sqrt{3} )=0\\cosx-1=0\\cosx=1\\x_1=2\pi n.\\2*sinx-\sqrt{3}=0\\ 2*sinx=\sqrt{3}\ |:2\\sinx=\frac{\sqrt{3} }{2} \\x_2=\frac{\pi }{3}+2\pi n\notin\\x_3=\frac{2\pi }{3}+2\pi n.$

Ответ: x₁=2πn, x₂=2π/3+2πn.