Решить уравнение:

[tex] \sqrt{2} \cdot ( 1 + \cos{x} + \sin{x} ) + 1 = \sqrt{3} \cdot \sin{x} - \cos{x} [/tex] ;


*** ответ не должен содержать в явном виде обратных функций:
[tex] arcsin(), arccos(), arctg() [/tex] или [tex] arcctg() . [/tex]

[tex]\sqrt{2}(1+cosx+sinx)+1=\sqrt{3}sinx-cosx [/tex]

[tex]\sqrt{2}+\sqrt{2}cosx+\sqrt{2}sinx+1=\sqrt{3}sinx-cosx[/tex]

Делим все на 2

[tex]\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cosx+\frac{\sqrt{2}}{2}sinx+\frac{1}{2}=\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx[/tex]

Преобразовываем

[tex]\frac{\sqrt{2}}{2}+sin(\frac{\pi}{4})cosx+sin(\frac{\pi}{4})sinx+\frac{1}{2}=cos(\frac{\pi}{6})sinx-sin(\frac{\pi}{6})cosx[/tex]

[tex]sin(\frac{\pi}{4})+sin(\frac{\pi}{4}+x)+sin(\frac{\pi}{6})=sin(x-\frac{\pi}{6})[/tex]

Группируем
 

[tex]sin(\frac{\pi}{4}+x)- sin(x-\frac{\pi}{6})= -sin(\frac{\pi}{4})- sin(\frac{\pi}{6})[/tex]

[tex]cos(\frac{\frac{\pi}{4}+x+x-\frac{\pi}{6}}{2})sin(\frac{\frac{\pi}{4}+x-x+\frac{\pi}{6}}{2})= -(sin(\frac{\pi}{4})+ sin(\frac{\pi}{6}))[/tex]

[tex]cos(x+\frac{\pi}{24})sin(\frac{5\pi}{24})= -sin(\frac{\frac{\pi}{4}+\frac{\pi}{6}}{2})cos\frac{\frac{\pi}{4}-\frac{\pi}{6}}{2}[/tex]

[tex]cos(x+\frac{\pi}{24})sin(\frac{5\pi}{24})= -sin(\frac{5\pi}{24})cos\frac{\pi}{24}[/tex]

[tex]cos(x+\frac{\pi}{24})= -cos\frac{\pi}{24}[/tex]

[tex]cos(x+\frac{\pi}{24})+cos\frac{\pi}{24}=0[/tex]

[tex]2cos( \frac{x+\frac{\pi}{24}+\frac{\pi}{24}}{2})cos(\frac{x+\frac{\pi}{24}-\frac{\pi}{24}}{2})=0[/tex]

[tex]2cos( \frac{x}{2}+\frac{\pi}{24})cos(\frac{x}{2}})=0[/tex]

[tex]cos( \frac{x}{2}+\frac{\pi}{24})cos(\frac{x}{2}})=0[/tex]

Получили два уравнения

[tex]cos( \frac{x}{2}+\frac{\pi}{24})=0[/tex]  и  [tex]cos(\frac{x}{2}})=0[/tex]

Из первого уравнения
[tex] \frac{x}{2}+\frac{\pi}{24}=\frac{\pi}{2}+ \pi*n[/tex]

[tex] \frac{x}{2}=\frac{\pi}{2}-\frac{\pi}{24}+ \pi*n[/tex]

[tex]\frac{x}{2}=\frac{11\pi}{24}+ \pi*n[/tex]

[tex]x=\frac{11\pi}{12}+ 2\pi*n[/tex]

Из второго уравнения

[tex]\frac{x}{2}=\frac{\pi}{2}+ \pi*n[/tex]

[tex]x=\pi+ 2\pi*n[/tex]