Question

ta có \(cos4x\)+ 2= 6\(sin^2x\)   (\(\dfrac{\Pi}{2}< x< \Pi\))    TÍnh \(tan2x\)

Answer 1

\(\dfrac{\pi}{2}< x< \pi\Rightarrow\pi< 2x< 2\pi\Rightarrow\left\{{}\begin{matrix}-1< cos2x< 1\\sin2x< 0\end{matrix}\right.\)

\(cos4x+2=6sin^2x\)

\(\Leftrightarrow2cos^22x-1+2=3\left(1-cos2x\right)\)

\(\Leftrightarrow2cos^22x+3cos2x+1=0\Rightarrow\left[{}\begin{matrix}cos2x=-1\left(loại\right)\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow sin2x=-\sqrt{1-cos^22x}=-\dfrac{\sqrt{3}}{2}\)

\(tan2x=\dfrac{sin2x}{cos2x}=\sqrt{3}\)