Question

Giải phương trình sau 

\(\dfrac{cos3x-cosx}{1+sinx}=0\)

Answer 1

ĐKXĐ: \(sinx\ne-1\)

=>\(x\ne-\dfrac{\Omega}{2}+k2\Omega\)

\(\dfrac{cos3x-cosx}{1+sinx}=0\)

=>cos3x-cosx=0

=>cos3x=cosx

=>\(\left[{}\begin{matrix}3x=x+k2\Omega\\3x=-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=k2\Omega\\4x=k2\Omega\end{matrix}\right.\Leftrightarrow x=\dfrac{k\Omega}{2}\)

mà \(x\ne-\dfrac{\Omega}{2}+k2\Omega\)

nên \(x=\dfrac{k\Omega}{2}\left(k\ne4n-1;n\in Z\right)\)