Question

2cos^22x+5sin2x-4=0

2cos^22x+2sinx=sin2x+2

Answer 1

Câu 1:

 \(\Leftrightarrow2\cdot\left(1-\sin^22x\right)+5\sin2x-4=0\)

\(\Leftrightarrow-2\sin^22x+5\sin2x-2=0\)

\(\Leftrightarrow\left(\sin2x-2\right)\left(2\sin2x-1\right)=0\)

\(\Leftrightarrow\sin2x=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{6}+k2\Pi\\2x=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{12}+k\Pi\\x=\dfrac{5}{12}\Pi+k\Pi\end{matrix}\right.\)