ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sinx}{cosx}+\frac{cosx}{sinx}=\sqrt{2}\left(sinx+cosx\right)\)
\(\Leftrightarrow\frac{1}{sinx.cosx}=\sqrt{2}\left(sinx+cosx\right)\Leftrightarrow\left(sinx+cosx\right)sinx.cosx=\frac{\sqrt{2}}{2}\)
Đặt \(sinx+cosx=a\) \(\left(\left|a\right|\le\sqrt{2}\right)\)
\(\Rightarrow a^2=1+2sinx.cox\Rightarrow sinx.cosx=\frac{a^2-1}{2}\) pt trở thành:
\(\left(a^2-1\right)a=\sqrt{2}\Leftrightarrow a^3-a-\sqrt{2}=0\)
\(\Leftrightarrow\left(a-\sqrt{2}\right)\left(a^2+a\sqrt{2}+1\right)=0\Rightarrow a=\sqrt{2}\)
\(\Rightarrow sinx+cosx=\sqrt{2}\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=1\)
\(\Rightarrow...\)