ĐKXĐ: \(x\ne k\dfrac{\pi}{2}\)
\(sinx+cosx+\dfrac{sinx+cosx}{sinx.cosx}=\dfrac{10}{3}\)
Đặt \(sinx+cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(sinx.cosx=\dfrac{t^2-1}{2}\)
Pt trở thành:
\(t+\dfrac{2t}{t^2-1}=\dfrac{10}{3}\)
\(\Rightarrow3t^3-10t^2+3t+10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2-4t-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{2-\sqrt{19}}{3}\\t=\dfrac{2+\sqrt{19}}{3}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow...\)