Question

2cos2x+1 / 2cosx-căn2 =0

Answer 1

ĐKXĐ: \(2\cdot cosx-\sqrt2<>0\)

=>\(2\cdot cosx<>\sqrt2\)

=>\(cosx<>\frac{\sqrt2}{2}\)

=>\(x\notin\left\lbrace\frac{\pi}{4}+k2\pi;-\frac{\pi}{4}+k2\pi\right\rbrace\)

\(\frac{2\cdot cos2x+1}{2\cdot cosx-\sqrt2}=0\)

=>\(2\cdot cos2x+1=0\)

=>\(cos2x=-\frac12\)

=>\(\left[\begin{array}{l}2x=\frac{2\pi}{3}+k2\pi\\ 2x=-\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{3}+k\pi\\ x=-\frac{\pi}{3}+k\pi\end{array}\right.\)