ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\Leftrightarrow3\left(\dfrac{cosx}{sinx}-cosx\right)-5\left(\dfrac{sinx}{cosx}-sinx\right)=2\)
\(\Leftrightarrow\dfrac{3\left(cosx-sinx.cosx\right)}{sinx}+3-\dfrac{5\left(sinx-sinx.cosx\right)}{cosx}-5=0\)
\(\Leftrightarrow\dfrac{3\left(sinx+cosx-sinx.cosx\right)}{sinx}-\dfrac{5\left(sinx+cosx-sinx.cosx\right)}{cosx}=0\)
\(\Leftrightarrow\left(sinx+cosx-sinx.cosx\right)\left(\dfrac{3}{sinx}-\dfrac{5}{cosx}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx-sinx.cosx=0\left(1\right)\\tanx=\dfrac{3}{5}\Rightarrow x=arctan\left(\dfrac{3}{5}\right)+k\pi\end{matrix}\right.\)
Xét (1), đặt \(sinx+cosx=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(sinx.cosx=t^2-1\)
\(\left(1\right)\Leftrightarrow t-\dfrac{t^2-1}{2}=0\Leftrightarrow t^2-2t-1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(loại\right)\\t=1-\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1-\sqrt{2}\)
\(\Rightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}-2}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arcsin\left(\dfrac{\sqrt{2}-2}{2}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arcsin\left(\dfrac{\sqrt{2}-2}{2}\right)+k2\pi\end{matrix}\right.\)
