Question

Tổng các nghiệm của phương trình \(\frac{1}{cosx}+\frac{1}{sin2x}=\frac{2}{sin4x}\) trên khoảng \(\left(0;\pi\right)\) là ?

Answer 1

ĐKXĐ:

\(\Leftrightarrow\frac{1}{cosx}+\frac{1}{sin2x}=\frac{1}{sin2x.cos2x}\)

\(\Leftrightarrow\frac{2sinx.cos2x}{sin2x.cos2x}+\frac{cos2x}{sin2x.cos2x}=\frac{1}{sin2x.cos2x}\)

\(\Leftrightarrow2sinx.cos2x+1-2sin^2x=1\)

\(\Leftrightarrow2sinx\left(cos2x-sinx\right)=0\)

\(\Leftrightarrow cos2x-sinx=0\)

\(\Leftrightarrow-2sin^2x-sinx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)

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