tan 2x-cotx=0 giải bài bên

Đáp án:

\(x=\dfrac{\pi}{6}+k.\dfrac{\pi}{3}\) \(k \epsilon Z; k \neq 1\)

Giải thích các bước giải:

ĐK: \(\sin x \neq 0\)

\(\Leftrightarrow x \neq k.\pi\)

 \(\tan 2x-\cot x=0\)

\(\Leftrightarrow \tan 2x=\tan(\dfrac{\pi}{2}-x)\)

\(\Leftrightarrow 2x=\dfrac{\pi}{2}-x+k.\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k.\dfrac{\pi}{3}\) \(k \epsilon Z\)

\(\dfrac{\pi}{6}+k.\dfrac{\pi}{3} \neq k.\pi\)

\(\Rightarrow k.\dfrac{\pi}{6} \neq \dfrac{\pi}{6}\)

\(\Rightarrow k \neq 1\)

Vậy \(x=\dfrac{\pi}{6}+k.\dfrac{\pi}{3}\) \(k \epsilon Z; k \neq 1\)