ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(sin2x+sinx-\dfrac{1}{2sinx}-\dfrac{1}{2sinx.cosx}=\dfrac{2cos2x}{2sinx.cosx}\)
\(\Rightarrow sin^22x+sin2x.sinx-cosx-1=2cos2x\)
\(\Leftrightarrow sin^22x-1+2sin^2x.cosx-cosx=2cos2x\)
\(\Leftrightarrow-cos^22x-cosx\left(1-2sin^2x\right)=2cos2x\)
\(\Leftrightarrow-cos^22x-cosx.cos2x=2cos2x\)
\(\Leftrightarrow cos2x\left(cos2x+cosx+2\right)=0\)
\(\Leftrightarrow cos2x\left(2cos^2x+cosx+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cos2x=0\\2cos^2x+cosx+1=0\left(vn\right)\\\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
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