Question

Giải phương trình:  \(\dfrac{1}{cosx}+\dfrac{1}{sin2x}=\dfrac{2}{sin4x}\)

 

Answer 1

ĐKXĐ: \(x\ne\dfrac{k\pi}{4}\)

\(\dfrac{1}{cosx}+\dfrac{1}{2sinx.cosx}=\dfrac{2}{4sinx.cosx.cos2x}\)

\(\Rightarrow2sinx.cos2x+cos2x=1\)

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

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

\(\Leftrightarrow2sinx\left(1-2sin^2x-sinx\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sinx=0\\sinx=-1\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\left(loại\right)\\x=-\dfrac{\pi}{2}+k2\pi\left(loại\right)\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)