\(\Leftrightarrow5\cdot\dfrac{1-cos2x}{2}-2sin2x+3\cdot\dfrac{1+cos2x}{2}=2\)
\(\Leftrightarrow\dfrac{5}{2}-\dfrac{5}{2}cos2x-2sin2x+\dfrac{3}{2}+\dfrac{3}{2}cos2x=2\)
\(\Leftrightarrow-2sin2x-cos2x=2-\dfrac{5}{2}-\dfrac{3}{2}=-2\)
=>2sin2x+cos2x=2
\(\Leftrightarrow\dfrac{2sin2x}{\sqrt{5}}+\dfrac{cos2x}{\sqrt{5}}=\dfrac{2}{\sqrt{5}}\)
\(\Leftrightarrow sin\left(2x+\alpha\right)=\dfrac{2}{\sqrt{5}}\), với cos a=2/căn 5
Đến đây thì dễ rồi
Đúng 3
Bình luận (0)
