Chủ đề này có 1 trả lời

[iamshant]

$2\sqrt{2}cos2x + sin2x. cos(x+\frac{3\pi }{4})-4sin(x+\frac{\pi }{4})=0$

[Atu]

$2\sqrt{2}cos2x + sin2x. cos(x+\frac{3\pi }{4})-4sin(x+\frac{\pi }{4})=0$

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}$ ....