Question

(2cosx-1)cotx = \(\frac{3}{sinx}\) + \(\frac{2sinx}{cosx-1}\)

Answer 1

ĐKXĐ: \(x\ne k\pi\)

\(\Leftrightarrow\frac{\left(2cosx-1\right).cosx}{sinx}-\frac{3}{sinx}=\frac{2sinx}{cosx-1}\)

\(\Leftrightarrow\frac{2cos^2x-cosx-3}{sinx}=\frac{2sinx}{cosx-1}\)

\(\Leftrightarrow\frac{\left(cosx+1\right)\left(2cosx-3\right)}{sinx}=\frac{2sinx}{cosx-1}\)

\(\Leftrightarrow\left(cos^2x-1\right)\left(2cosx-3\right)=2sin^2x\)

\(\Leftrightarrow-sin^2x\left(2cosx-3\right)=2sin^2x\)

\(\Leftrightarrow2cosx-3=-2\Rightarrow cosx=\frac{1}{2}\)

\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)