$2\sqrt{2}cos2x + sin2x. cos(x+\frac{3\pi }{4})-4sin(x+\frac{\pi }{4})=0$
Ta có:
$2\sqrt{2}cos2x + sin2x. cos(x+\frac{3\pi }{4})-4sin(x+\frac{\pi }{4})=0$
$\Leftrightarrow 2\sqrt{2}cos2x+sin2x.cos(x+\pi-\frac{\pi }{4} )-2\sqrt{2}(sinx+cosx)=0$
$\Leftrightarrow 2\sqrt{2}cos2x-sin2x.cos(x-\frac{\pi }{4})2\sqrt{2}(sinx+cosx)=0$
$\Leftrightarrow 2\sqrt{2}(sinx+cosx)(cosx-sinx)-\sqrt{2}sinx.cosx.(sinx+cosx)-2\sqrt{2}(sinx+cosx)=0$
$\Leftrightarrow (sinx+cosx)(2cosx-2sinx-sinx.cosx-2)=0$
Đặt $ sinx-cosx=t\Leftrightarrow -sinxcosx=\frac{t^{2}-1}{2}$ ....