2cos(x-pi/4)=(√2-2 sinx)sinx

[latex]2\cos(x- \frac{\pi}{4} )=( \sqrt{2} -2\sin x)\sin x \\ 2(\cos x\cos \frac{\pi}{4}+\sin x\sin\frac{\pi}{4})=\sqrt{2}\sin x-2\sin^2x \\ \sqrt{2}\cos x+\sqrt{2}\sin x=\sqrt{2}\sin x-2\sin^2x\\ 2\sin^2x+\sqrt{2}\cos x=0 \\ 2-2\cos^2x+\sqrt{2}\cos x=0 \\ 2\cos^2x-\sqrt{2}\cos x-2=0[/latex] пусть cos x = t (|t|≤1), тогда получаем [latex]2t^2-\sqrt{2}t-2=0 \\ D=b^2-4ac=(-\sqrt{2})^2-4\cdot2\cdot (-2)=18;\,\,\, \sqrt{D} =3\sqrt{2} \\ t_1= \frac{\sqrt{2}-3\sqrt{2}}{4} =- \frac{\sqrt{2}}{2} \\ t_2= \frac{\sqrt{2}+3\sqrt{2}}{4} =\sqrt{2}[/latex] Корень t=√2 не удовлетворяет условие при |t|≤1 Обратная замена [latex]\cos x=- \frac{\sqrt{2}}{2} \\ x=\pm \frac{3 \pi }{4} +2 \pi n,n \in Z[/latex]