Question

Giải phương trình 

\(\dfrac{3}{cosx}+tan^2x=9\)

Answer 1

\(\dfrac{3}{cosx}+tan^2x=9\)
\(\Leftrightarrow\dfrac{sin^2x}{cos^2x}+\dfrac{3}{cosx}=9\)
\(\Leftrightarrow9cos^2x-3cosx-sin^2x=0\)
\(\Leftrightarrow9cos^2x-3cosx-1+cos^2x=0\)
\(\Leftrightarrow10cos^2x-3cosx-1=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(5cosx+1\right)=0\)
\(\Leftrightarrow cosx=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}cosx=cos\dfrac{\pi}{3}\\cosx=-\dfrac{1}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=arcsin\left(-\dfrac{1}{5}\right)+k2\pi\end{matrix}\right.\)