Question

Giải pt sau: \(\sin^2x+\sin2x-2\cos^2x=\dfrac{1}{2}\)

Answer 1

Pt \(\Leftrightarrow sin^2x+2.sinx.cosx-2cos^2x=\dfrac{1}{2}\left(sin^2x+cos^2x\right)\)

\(\Leftrightarrow sin^2x.\dfrac{1}{2}+2.sinx.cosx-\dfrac{5}{2}cos^2x=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+5cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\sinx=-5cosx\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-5\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arc.tan\left(-5\right)+k\pi\end{matrix}\right.\)(\(k\in Z\))

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