Chọn A.
\(2sin^2x-cosx+1=0\Rightarrow2\cdot\dfrac{1-cos2x}{2}-cosx+1=0\)
\(\Rightarrow1-\left(2cos^2x-1\right)-cosx+1=0\)
\(\Rightarrow-2cos^2x-cosx+3=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\)\(\Rightarrow x=k2\pi\)
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