Срочно!!! Решите уравнение.cos (П+х) + sin П+х/2 = 1и найти значения на промежутке (3П; 9П/2]

[latex]cos(\pi +x)+sin(\pi +\frac{x}{2})=1\\\\-cosx-sin\frac{x}{2}=1\\\\-(cos^2\frac{x}{2}-sin^2\frac{x}{2})-sin\frac{x}{2}=1\\\\sin^2\frac{x}{2}-cos^2\frac{x}{2}-sin\frac{x}{2}=1\\\\sin^2\frac{x}{2}-(1-sin^2\frac{x}{2})-sin\frac{x}{2}=1\\\\2sin^2\frac{x}{2}-sin\frac{x}{2}-2=0\\\\t=sin\frac{x}{2},\; 2t^2-t-2=0\\\\D=1+16=17[/latex] [latex]2)))cos(\pi+x)+sin(\frac{\pi}{2}+\frac{x}{2})=1\\\\-cosx+cos\frac{x}{2} =1\\\\2cos^2\frac{x}{2}-cos\frac{x}{2}=0\\\\cos\frac{x}{2}(2cos\frac{x}{2}-1)=0\\\\cos\frac{x}{2}=0,\frac{x}{2}=2\pi n ,x=4\pi n\\\\cos\frac{x}{2}=\frac{1}{2},\frac{x}{2}=\pm\frac{\pi}{3}+2\pin\\\\x=\pm \frac{2\pi}{3}+4\pi n\\\\x_1=4\pi ,x=4\pi -\frac{2\pi}{3}=\frac{10\pi}{3})))))))[/latex] [latex]t_1=\frac{1-\sqrt{17}}{4}=-1,t_2=\frac{1+\sqrt{17}}{4}[/latex] [latex]sin\frac{x}{2}=\frac{1+\sqrt{17}}{4}>1\; net\; resheniya\; (\sqrt{17}\approx 4,12)\\\\sin\frac{x}{2}=\frac{1-\sqrt{17}}{4},\frac{x}{2}=(-1)^{n}arcsin\frac{1-\sqrt{17}}{4}+\pi n,\; n\in Z\\\\x=2arcsin\frac{1-\sqrt{17}}{4}+2\pi n\\\\x=4\pi -arcsin\frac{1-\sqrt{17}}{4}\in (3\pi ;\frac{9\pi}{2}][/latex] [latex]x=3\pi +arcsin\frac{1-\sqrt{17}}{4}}\in (3\pi;\frac{9\pi}{2}][/latex]