Тригонометрия. Решите уравнение cos2x=sin(3п/2-x); [3п/2;5п/2].

[latex]cos2x=sin(\frac{3\pi }{2}-x)\\\\cos2x=-cosx\\\\cos^2x-sin^2x=-cosx\\\\cos^2x-(1-cos^2x)+cosx=0\\\\2cos^2x+cosx-1=0\\\\t=cosx,\; 2t^2+t-1=0\\\\D=1+8=9,\; t_1=\frac{-1-3}{4}=-1,\; t_2=\frac{1}{2}\\\\cosx=-1,\; x=\pi +2\pi n,\; n\in Z\\\\cosx=\frac{1}{2},x=\pm \frac{\pi }{3}+2\pi k,\; k\in Z[/latex] [latex]x_1=-\frac{\pi}{3}+2\pi =\frac{5\pi }{3},x_2=\frac{\pi}{3}+2\pi =\frac{7\pi }{3}\; \in [\frac{3\pi }{2},\frac{5\pi }{2}][/latex]