Question

В ответ надо записать сумму корней в градусах, которые принадлежат промежутку [0°;180°]

Answer 1

$\displaystyle (cosx-sinx)^2=\frac{3\sqrt{3}-3}{2\sqrt{3}}(1-sin^22x)\\\\(cos^2x-2sinx*cosx+sin^2x)=\frac{3-\sqrt{3}}{2}(1-sin^22x)\\\\1-sin2x=\frac{1}{2}((3-\sqrt{3})-(3-\sqrt{3})sin^22x)\\\\2-2sin2x-(3-\sqrt{3})+(3-\sqrt{3})sin^22x=0\\\\sin2x=t; |t|\leq 1\\\\(3-\sqrt{3})t^2-2t+(2-3+\sqrt{3})=0\\\\(3-\sqrt{3})t^2-2t+(\sqrt{3}-1)=0\\\\D=4-4*(3-\sqrt{3})(\sqrt{3}-1)=4(1-(3\sqrt{3}-3-3+\sqrt{3}))= \\\\=4(1-4\sqrt{3}+6 )=4(7-4\sqrt{3})=(2(\sqrt{3}-2))^2\\\\t_{1.2}=\frac{2 \pm 2(\sqrt{3}-2)}{2(3-\sqrt{3})}$

$\displaystyle t_{1.2}=\frac{1 \pm(\sqrt{3}-2)}{3-\sqrt{3}}\\\\t_1=\frac{1-\sqrt{3}+2}{3-\sqrt{3}}=\frac{3-\sqrt{3}}{3-\sqrt{3}}=1\\\\t_2=\frac{1+\sqrt{3}-2}{3-\sqrt{3}}=\frac{(\sqrt{3}-1)(3+\sqrt{3})}{(3-\sqrt{3})(3+\sqrt{3})}=\frac{3\sqrt{3}-3+3-\sqrt{3}}{9-3}=\frac{2\sqrt{3}}{6}=\frac{\sqrt{3}}{3}$

$\displaystyle sin2x=1; 2x=\frac{\pi }{2}+2\pi n; n \in Z; x=\frac{\pi }{4}+\pi n; n\in Z \\\\sin2x=\frac{1}{\sqrt{3}}; 2x=\left \{ {{arcsin\frac{1}{\sqrt{3}}+2\pi n; n \in Z} \atop {\pi -arcsin\frac{1}{\sqrt{3}}+2\pi n; n \in Z}} \right.; x=\left \{ {{\frac{1}{2}arcsin\frac{1}{\sqrt{3}}+\pi n; n\in Z} \atop {\frac{1}{2}(\pi -arcsin\frac{1}{\sqrt{3}})+\pi n; n\in Z}} \right.$