Решить уравнение!cos2x=4корня из 5 sin(x-пи/2)-7

[latex]cos(2x)=4\sqrt{5}sin(x-\frac{\pi}{2})-7[/latex] [latex]2cos^2x-1=-4\sqrt{5}cos x-7[/latex] [latex]cos x=t; -1 \leq t \leq 1[/latex] [latex]2t^2-1=-4\sqrt{5}t-7[/latex] [latex]2t^2+4\sqrt{5}t+6=0[/latex] [latex]t^2+2\sqrt{5}+3=0[/latex] [latex]D=(2\sqrt{5})^2-4*1*3=4*5-12=8=(2\sqrt{2})^2[/latex] [latex]t_1=\frac{-2\sqrt{5}-2\sqrt{2}}{2}=-\sqrt{5}-\sqrt{2}<-1[/latex] [latex]t_2=\frac{-2\sqrt{5}+2\sqrt{2}}{2}=\sqrt{2}-\sqrt{5}[/latex] [latex]cos x=\sqrt{2}-\sqrt{5}[/latex] [latex]x=^+_-arccos(\sqrt{2}-\sqrt{5})+2\pi*k[/latex] k є Z