ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(\sqrt{3}sin2x+cos^2x\left(\dfrac{sin^2x-cos^2x}{cos^2x}\right)=2\)
\(\Rightarrow\sqrt{3}sin2x-cos2x=2\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=1\)
\(\Leftrightarrow2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)
Đúng 3
Bình luận (1)