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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) \(\Leftrightarrow sin3x-cos3x=0\) \(\Leftrightarrow\sqrt{2}sin\left(3x-\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow3x-\dfrac{\pi}{4}=k\pi\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\left(k\in Z\right)\)
Vậy ...
Với \(cosx=0\) ko phải nghiệm
Với \(cosx\ne0\)
\(\Rightarrow\left(2sin5x-1\right)\left(2cos2x.cosx-cosx\right)=2sinx.cosx\)
\(\Leftrightarrow\left(2sin5x-1\right)\left(cos3x+cosx-cosx\right)=sin2x\)
\(\Leftrightarrow cos3x\left(2sin5x-1\right)=sin2x\)
\(\Leftrightarrow2sin5x.cos3x-cos3x=sin2x\)
\(\Leftrightarrow sin8x+sin2x-cos3x=sin2x\)
\(\Leftrightarrow sin8x=cos3x=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow...\)
=>\(2\cdot cos2x\cdot sin2x+2cos^22x-sin2x-cos2x-1=0\)
=>\(2cos2x\cdot sin2x+2\cdot cos^22x-1=sin2x+cos2x\)
=>\(sin4x+cos4x=sin2x+cos2x\)
=>\(sin\left(4x+\dfrac{pi}{4}\right)=sin\left(2x+\dfrac{pi}{4}\right)\)
=>4x+pi/4=2x+pi/4+k2pi hoặc 4x+pi/4=pi-2x-pi/4+k2pi
=>2x=k2pi hoặc 6x=1/2pi+k2pi
=>x=kpi hoặc x=1/12pi+kpi/3
\(\left(2cos2x-1\right)\left(sin2x+cos2x\right)=1\)
\(\Leftrightarrow2sin2x.cos2x+2cos^22x-sin2x-cos2x-1=0\)
\(\Leftrightarrow sin4x+cos4x-sin2x-cos2x=0\)
\(\Leftrightarrow2cos3x.sinx-2sin3x.sinx=0\)
\(\Leftrightarrow2sinx\left(cos3x-sin3x\right)=0\)
\(\Leftrightarrow2\sqrt{2}sinx.cos\left(3x+\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos\left(3x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\3x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\end{matrix}\right.\)
Nhận thấy \(sinx=0\) ko phải nghiệm
\(\Leftrightarrow sinx\left(2cos2x+1\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)
\(\Leftrightarrow\left(2cos2x.sinx+sinx\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)
\(\Leftrightarrow\left(sin3x-sinx+sinx\right)\left(2cos6x+1\right)\left(2cos18x+1\right)=sinx\)
\(\Leftrightarrow\left(2cos6x.sin3x+sin3x\right)\left(2cos18x+1\right)=sinx\)
\(\Leftrightarrow\left(2cos18x.sin9x+sin9x\right)=sinx\)
\(\Leftrightarrow sin27x=sinx\)