Question

Giải pt (2cos2x-1)(sin2x+cos2x)=1

Answer 1

P/t \(\Leftrightarrow2cos2x.sin2x-sin2x+2cos^22x-cos2x-1=0\)

\(\Leftrightarrow sin4x-sin2x+cos4x-cos2x=0\)

\(\Leftrightarrow2sinx.cos3x-2sin3x.sinx=0\)

\(\Leftrightarrow sinx\left(cos3x-sin3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(1\right)\\cos3x=sin3x\left(2\right)\end{matrix}\right.\) 

(1) \(\Leftrightarrow x=k\pi\left(k\in Z\right)\)

(2) \(\Leftrightarrow sin3x-cos3x=0\)  \(\Leftrightarrow\sqrt{2}sin\left(3x-\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow3x-\dfrac{\pi}{4}=k\pi\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\left(k\in Z\right)\)

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