Question

тригонометрия
пожалуйста подробно

Answer 1

$\displaystyle sin2x=\sqrt{2}sin(\frac{3\pi }{2}-x)\\\\2sinx*cosx=\sqrt{2}(-cosx)\\\\cosx(2sinx+\sqrt{2})=0\\\\cosx=0; ili : sinx=-\frac{\sqrt{2}}{2}\\\\cosx=0; x=\frac{\pi }{2}+\pi n; n \in Z\\\\sinx=-\frac{\sqrt{2}}{2}; x_1=-\frac{\pi }{4}+2\pi k;k\in Z; x_2=-\frac{3\pi }{4}+2\pi k; k\in Z$

отбор корней [π;5π/2]

5π/4; 3π/2; 7π/4; 5π/2