4 - sinxcosx + 4cosx = 4sinx

[latex]4-\sin x\cos x+4\cos x=4\sin x \\ \\ 4(\cos^2x+\sin^2x)-\sin x\cos x+4\cos x-4\sin x=0 \\ 4(\cos^2+\sin^2-2\sin x\cos x+2\sin x\cos x)-\sin x\cos x+4(\cos x-\sin x)=0[/latex] [latex]4((\cos x-\sin x)^2+2\sin x\cos x)-\sin x\cos x+4(\cos x-\sin x)=0 \\ 4(\cos x-\sin x)^2+7\sin x\cos x+4(\cos x-\sin x)=0[/latex]  Пусть [latex](\sin x-\cos x)=t(|t| \leq \sqrt{2} )[/latex], тогда [latex]1-2\sin x\cos x=t^2 \\ \sin x\cos x= \frac{1-t^2}{2} [/latex] Получаем [latex]4t^2+7 \frac{1-t^2}{2} +4t=0 \\ 8t^2+7-7t^2+8t=0 \\ t^2+8t+7=0 \\ D=64-28=36 \\ t_1=-1 \\ t_2=-7[/latex] Корень t=-7 - не удовлетворяет условие при |t|≤√2 Возвращаемся к замене [latex]\cos x-\sin x=-1 \\ \sqrt{2} \sin (x- \frac{ \pi }{4} )=-1 \\ x-\frac{ \pi }{4}=(-1)^k^+^1\cdot \frac{ \pi }{4}+ \pi k,k \in Z \\ x=(-1)^k^+^1\cdot \frac{ \pi }{4}+\frac{ \pi }{4}+ \pi k,k \in Z[/latex]