Question

\(cos2x+2cosx-sin^2\dfrac{x}{2}=0\)

Giải phương trình lượng giác

Answer 1

\(\Leftrightarrow2cos^2x-1+2cosx-\left(\dfrac{1}{2}-\dfrac{1}{2}cosx\right)=0\)

\(\Leftrightarrow2cos^2x+\dfrac{5}{2}cosx-\dfrac{3}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{-5+\sqrt{73}}{8}\\cosx=\dfrac{-5-\sqrt{73}}{8}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm arccos\left(\dfrac{-5+\sqrt{73}}{8}\right)+k2\pi\)