\(\Leftrightarrow2\left(sin^2x+cos^2x\right)^2-4sin^2x.cos^2x+sin\left(4x-\dfrac{\pi}{2}\right)+sin2x-3=0\)
\(\Leftrightarrow2-sin^22x-cos4x+sin2x-3=0\)
\(\Leftrightarrow-sin^22x+2sin^22x-1+sin2x-1=0\)
\(\Leftrightarrow sin^22x+sin2x-2=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-2\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{\pi}{4}+k\pi\)
Đúng 1
Bình luận (0)
