ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow cosx+sinx+\frac{cosx+sinx}{sinx.cosx}=\frac{10}{3}\)
\(\Leftrightarrow sinx.cosx\left(sinx+cosx\right)+sinx+cosx=\frac{10}{3}sinx.cosx\)
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow\frac{t\left(t^2-1\right)}{2}+t=\frac{5}{3}\left(t^2-1\right)\)
\(\Leftrightarrow3t^3-10t^2+3t+10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2-4t-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(l\right)\\t=\frac{2+\sqrt{19}}{3}\left(l\right)\\t=\frac{2-\sqrt{19}}{3}\end{matrix}\right.\) \(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\frac{2-\sqrt{19}}{3}\)
\(\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{2\sqrt{2}-\sqrt{38}}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=arcsin\left(\frac{2\sqrt{2}-\sqrt{38}}{6}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(\frac{2\sqrt{2}-\sqrt{38}}{6}\right)+k2\pi\end{matrix}\right.\)
