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NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 8 2020
\(\Leftrightarrow sin4x\left(sin5x+sin3x\right)-sin2x.sinx=0\)
\(\Leftrightarrow2sin^24x.cosx-2sin^2x.cosx=0\)
\(\Leftrightarrow cosx\left(2sin^24x-2sin^2x\right)=0\)
\(\Leftrightarrow cosx\left(1-cos8x-1+cos2x\right)=0\)
\(\Leftrightarrow cosx\left(cos2x-cos8x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos8x=cos2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\8x=2x+k2\pi\\8x=-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{k\pi}{3}\\x=\frac{k\pi}{5}\end{matrix}\right.\)
TD
17
NT
1
21 tháng 7 2022
a: \(\Leftrightarrow2\cdot\sin3x\cdot\cos x-2\cos^2x=0\)
\(\Leftrightarrow\cos x\left(\sin3x-\cos x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\\sin3x=\cos x=\sin\left(\dfrac{\Pi}{2}-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\3x=\dfrac{\Pi}{2}-x+k2\Pi\\3x=\dfrac{\Pi}{2}+x+k2\Pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{2}\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)
b: \(\Leftrightarrow\sin x+\sin5x+\sin^2x=0\)
\(\Leftrightarrow\sin x=0\)
hay \(x=k\Pi\)
4 tháng 9 2018
sin3x + 1=2sin22x
<=> sin3x + 1 = 2\(\dfrac{1-cos4x}{2}\)
<=> sin3x + 1 = 1 - cos4x
<=> sin3x = -cos4x
<=> sin3x + cos4x = 0
<=> \(\dfrac{\sqrt{2}}{2}\)sin3x + \(\dfrac{\sqrt{2}}{2}\)cos4x = 0 (chia 2 vế cho \(\sqrt{2}\)).
<=> cos\(\dfrac{\pi}{4}\)sin3x + sin\(\dfrac{\pi}{4}\)cos4x = 0
<=> sin (3x+\(\dfrac{\pi}{4}\)) = 0
<=> sin(3x+\(\dfrac{\pi}{4}\)) = sin0
<=> \(\left[{}\begin{matrix}3x+\dfrac{\pi}{4}=0+k2\pi\\3x+\dfrac{\pi}{4}=\pi-0+k2\pi\end{matrix}\right.\)(k\(\in\)Z)
<=>\(\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+\dfrac{k2\pi}{3}\\x=\dfrac{5\pi}{12}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(k\(\in\)Z)
\(\sin\left(5x\right)+\sin\left(3x\right)+2\cos\left(x\right)=1+\sin\left(4x\right)\)
\(\Leftrightarrow2\sin\left(4x\right)\cos\left(x\right)-\sin\left(4x\right)+2\cos\left(x\right)-1=0\)
\(\Leftrightarrow\sin\left(4x\right)(2\cos\left(x\right)-1)+(2\cos\left(x\right)-1)=0\)
\(\Leftrightarrow(2\cos\left(x\right)-1)(\sin\left(4x\right)+1)=0\)
\(\Rightarrow\left[{}\begin{matrix}\cos\left(x\right)=\dfrac{1}{2}\\\sin\left(4x\right)=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{-\pi}{2}+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{-\pi}{8}+k\dfrac{\pi}{2}\end{matrix}\right.\)