ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne-\dfrac{1}{2}\\sinx\ne1\end{matrix}\right.\)
\(\left(1-2sinx\right)cosx=\sqrt{3}\left(1+2sinx\right)\left(1-sinx\right)\)
\(\Leftrightarrow cosx-2sinx.cosx=\sqrt{3}+\sqrt{3}sinx-2\sqrt{3}sin^2x\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}+\sqrt{3}sinx-\sqrt{3}\left(1-cos2x\right)\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}sinx+\sqrt{3}cos2x\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x+\sqrt{3}cos2x\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow...\)
Đúng 2
Bình luận (0)
