​

ответ:

разделим на 2 каждый член уравнения

\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cos x =\frac{\sqrt{2}}{2}

2

3

sinx+

2

1

cosx=

2

2

\begin{lgathered}\frac{\sqrt{3}}{2}=cos{\frac{\pi}{6}}\\ \frac{1}{2}=sin{\frac{\pi}{6}}\\ sin(x+\frac{\pi}{6})=\frac{\sqrt{2}}{2}\\ x+\frac{\pi}{6} = \frac{\pi}{4}+2\pi n\\ x= -\frac{\pi}{6} + \frac{\pi}{4}+2\pi n\\ x = \frac{\pi}{12}+2\pi n\\ \\ x+\frac{\pi}{6} = \pi-\frac{\pi}{4}+2\pi n\\ x+\frac{\pi}{6} = \frac{3\pi}{4}+2\pi n\\ x=-\frac{\pi}{6} + \frac{3\pi}{4}+2\pi n\\ x = \frac{7\pi}{12}+2\pi {lgathered}

2

3

=cos

6

π

2

1

=sin

6

π

sin(x+

6

π

)=

2

2

x+

6

π

=

4

π

+2πn

x=−

6

π

+

4

π

+2πn

x=

12

π

+2πn

x+

6

π

=π−

4

π

+2πn

x+

6

π

=

4

3π

+2πn

x=−

6

π

+

4

3π

+2πn

x=

12

7π

+2πn