\Leftrightarrow[TEX]2\sqrt{2}sin(x-\frac{\pi}{3}+\frac{\pi}{4})cosx=1[/TEX]
\Leftrightarrow[TEX]2\sqrt{2}cosx[\frac{1}{\sqrt{2}}sin(x-\frac{\pi}{3})+\frac{1}{\sqrt{2}}cos(x-\frac{\pi}{3})=1[/TEX]
\Leftrightarrow[TEX]2cosx(sinx \frac{1}{2}-\frac{\sqrt{3}}{2}cos +cosx\frac{1}{2}+\frac{\sqrt{3}}{2}sinx)=1[/TEX]
[tex]\Leftrightarrow cosx(sinx+cosx+\sqrt{3}sinx-\sqrt{3}cosx)=1[/tex]
[TEX]\Leftrightarrow sinxcosx+cos^2x+\sqrt{3}sinxcosx -\sqrt{3}cos^2x=1 \\\\ \Leftrightarrow \frac{1+\sqrt{3}}{2}sin2x +\frac{1-\sqrt{3}}{2}(2cos^2x-1)-\frac{1+\sqrt{3}}{2}=0 \\\\ \Leftrightarrow \frac{1+\sqrt{3}}{2}sin2x +\frac{1-\sqrt{3}}{2}cos2x-\frac{1+\sqrt{3}}{2}=0\\\\ \Leftrightarrow (1+\sqrt{3})sin2x +(1-\sqrt{3})cos2x-(1+\sqrt{3})=0(1) \\\\ x=\frac{\pi}{2}+k\pi : \ la \ nghiem \ hay k ? \\\\ voi \ x \ khac \ \frac{\pi}{2}+k\pi ---> dat: tanx=t\\\\ (1) \Leftrightarrow (1+\sqrt{3})\frac{2t}{1+t^2}+(1-\sqrt{3})\frac{1-t^2}{1+t^2}-(1+\sqrt{3})=0 \\\\ \Leftrightarrow t^2-(1+\sqrt{3})t+\sqrt{3}=0 \\\\ \left[\begin{ t=1 \\ t=\sqrt{3} \right. \\\\ \Leftrightarrow \left[\begin{ x=\frac{\pi}{4}+k\pi \\ x= \frac{\pi}{3}+k\pi[/TEX]
có thể giải (1) theo dạng : asinx+bcosx=c