Question

\(\dfrac{tan^23x-tan^2x}{1-tan^23x.tan^2x}=1\) có nghiệm là ?

Answer 1

Làm lại:

ĐK: \(x\ne\dfrac{\pi}{2}+k\pi;x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2};x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{3}\)

\(\dfrac{tan^23x-tan^2x}{1-tan^23x.tan^2x}=1\)

\(\Leftrightarrow\dfrac{tan3x-tanx}{1+tan3x.tanx}.\dfrac{tan3x+tanx}{1-tan3x.tanx}=1\)

\(\Leftrightarrow tan2x.tan4x=1\)

\(\Leftrightarrow\dfrac{sin2x.sin4x}{cos2x.cos4x}=1\)

\(\Leftrightarrow sin2x.sin4x=cos2x.cos4x\)

\(\Leftrightarrow\dfrac{1}{2}\left(cos2x-cos6x\right)=\dfrac{1}{2}\left(cos6x+cos2x\right)\)

\(\Leftrightarrow cos6x=0\)

\(\Leftrightarrow6x=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{6}\)

Đối chiếu với điều kiện rồi kết luận.