Question

\(5sinx-2=3\left(1-sinx\right)tan^2x\)

Answer 1

ĐKXĐ: \(cosx\ne0\)

Đặt \(sinx=t\)

\(\Rightarrow5t-2=3\left(1-t\right).\frac{t^2}{1-t^2}\)

\(\Leftrightarrow5t-2=\frac{3t^2}{1+t}\)

\(\Leftrightarrow\left(5t-2\right)\left(1+t\right)=3t^2\)

\(\Leftrightarrow2t^2+3t-2=0\Rightarrow\left[{}\begin{matrix}t=\frac{1}{2}\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)