Задача 54624 решите способом введения дополнительного...
Условие
Решение
Делим на sqrt(2):
[m]\frac{1}{\sqrt{2}}sinx-\frac{1}{\sqrt{2}}cosx=\frac{\sqrt{3}}{2}[/m]
Так как
[m]\frac{1}{\sqrt{2}}=sin\frac{\pi}{4}[/m]
[m]\frac{1}{\sqrt{2}}=cos\frac{\pi}{4}[/m]
получаем уравнение:
[m]sin\frac{\pi}{4}\cdot sinx -cos\frac{\pi}{4}\cdot cosx=\frac{\sqrt{3}}{2}[/m]
Применяем формулу
cos α cos β -sin α sin β =cos( α + β )
[m] -cos(x+\frac{\pi}{4})=\frac{\sqrt{3}}{2}[/m]
[m] cos(x+\frac{\pi}{4})=-\frac{\sqrt{3}}{2}[/m]
[m] x+\frac{\pi}{4}=\pm arccos(-\frac{\sqrt{3}}{2})+2\pi n, n \in Z[/m]
[m] x+\frac{\pi}{4}=\pm (\pi -\frac{\pi}{6})+2\pi n, n \in Z[/m]
[m] x=-\frac{\pi}{4}\pm (\pi -\frac{\pi}{6})+2\pi n, n \in Z[/m] - о т в е т.
Аналогично
2)
Делим на sqrt(2):
[m]\frac{1}{\sqrt{2}}sin2x+\frac{1}{\sqrt{2}}cos2x=-\frac{1}{\sqrt{2}}[/m]
[m]sin\frac{\pi}{4}\cdot sin2x +cos\frac{\pi}{4}\cdot cos2x=-\frac{1}{\sqrt{2}}[/m]
[m] cos(2x-\frac{\pi}{4})=-\frac{1}{\sqrt{2}}[/m]
[m] 2x-\frac{\pi}{4}=\pm arccos(-\frac{1}{\sqrt{2}})+2\pi n, n \in Z[/m]
[m] 2x-\frac{\pi}{4}=\pm (\pi -\frac{\pi}{4})+2\pi n, n \in Z[/m]
[m] 2x=\frac{\pi}{4}\pm (\frac{3\pi}{4})+2\pi n, n \in Z[/m]
[m] x=\frac{\pi}{8}\pm (\frac{3\pi}{8})+\pi n, n \in Z[/m]- о т в е т.
3) Делим на 2:
[m]\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx=1[/m]
[m] cos(x-\frac{\pi}{4})=1[/m]
[m] x-\frac{\pi}{4}=2\pi n, n \in Z[/m]
[m] x=\frac{\pi}{4}+2\pi n, n \in Z[/m]- о т в е т.
4)Делим на 2:
[m]\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=\frac{\sqrt{3}}{2}[/m]
[m]sin\frac{\pi}{3}\cdot sin2x -cos\frac{\pi}{3}\cdot cos2x=\frac{\sqrt{3}}{2}[/m]
[m] -cos(2x+\frac{\pi}{3})=\frac{\sqrt{3}}{2}[/m]
[m] cos(2x+\frac{\pi}{3})=-\frac{\sqrt{3}}{2}[/m]
[m] cos(2x+\frac{\pi}{3})=\pm arccos(-\frac{\sqrt{3}}{2})+2\pi n, n \in Z[/m]
[m] 2x+\frac{\pi}{3}=\pm (\pi -\frac{\pi}{6})+2\pi n, n \in Z[/m]
[m] 2x=-\frac{\pi}{3}\pm (\frac{5\pi}{6})+2\pi n, n \in Z[/m]
[m] x=-\frac{\pi}{6}\pm (\frac{5\pi}{12})+\pi n, n \in Z[/m] - о т в е т.