Question

Найти наименьший корень тригонометрического уравнения.
$\alpha_{0}$ - наименьший на промежутке [π;2π] корень уравнения
$cos3x+sin(9x+2)=0$, тогда $sin(2 \alpha_{0} + \frac{1}{3} )$ равен
1) 1 2) $\frac{ \sqrt{2} }{2}$ 3) 0 4) $-\frac{ \sqrt{2} }{2}$ 5) -1

Answer 1

$\sin (\frac{\pi}{2}+3x) + \sin (9x+2)=0 \\ \\ 2 \sin (\frac{\frac{\pi}{2}+3x+9x+2}{2}) \cdot \cos (\frac{\frac{\pi}{2}+3x - 9x -2}{2})=0 \\ \\ 2 \sin (\frac{\pi}{4}+6x+1) \cdot \cos (\frac{\pi}{4}-3x -1)=0 \\ \\ \sin (6x+\frac{\pi}{4}+1) \cdot \cos (3x - \frac{\pi}{4} +1)=0 \\ \\ \sin (6x+\frac{\pi}{4}+1)=0; \ \ \ \ \ \ \ \ \ \ \ \ \cos (3x - \frac{\pi}{4} +1)=0 \\ \\ 6x+\frac{\pi}{4}+1=\pi n , \ n \in Z; \ \ \ \ \ 3x - \frac{\pi}{4} +1=\frac{\pi}{2}+\pi k, \ k \in Z$


$\\ \\ 6x=-\frac{\pi}{4}-1 + \pi n , \ n \in Z; \ \ \ \ \ 3x =\frac{3\pi}{4} -1+\pi k, \ k \in Z \\ \\ x=-\frac{\pi}{24} - \frac{1}{6} + \frac{\pi n}{6} , \ n \in Z; \ \ \ \ \ \ \ \ \ x =\frac{\pi}{4}- \frac{1}{3}+\frac{\pi k}{3}, \ k \in Z \\ \\ n=7: \ \ -\frac{\pi}{24} + \frac{7\pi}{6} - \frac{1}{6} =\frac{27\pi}{24}-\frac{1}{6}=\frac{1}{24} \cdot (27 \pi-4) \\ \\ k=3 : \ \ \frac{\pi}{4}+\pi -\frac{1}{3}=\frac{5 \pi}{4}- \frac{1}{3}=\frac{1}{12} \cdot (15 \pi -4)=\frac{1}{24} \cdot (30 \pi -8)$


$\boxed{\frac{1}{24} \cdot (27 \pi-4) } \ \textless \ \frac{1}{24} \cdot (30 \pi -8) \\ \\ \sin(2 \cdot (\frac{1}{24} \cdot (27 \pi-4) )+\frac{1}{3}) =\sin(\frac{27 \pi}{12} -\frac{4}{12}+\frac{1}{3}) =\sin(\frac{9 \pi}{4} -\frac{1}{3}+\frac{1}{3})=\\ \\ = \sin\frac{9 \pi}{4} = \sin(2 \pi+\frac{\pi}{4})=\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$


Ответ: 2)