sinx + sin2x + sin3x = 0
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26 tháng 7 2019
\( 2)\sin x + \sin 2x + \sin 3x = 0\\ \Leftrightarrow 2\sin 2x.\cos x + \sin 2x = 0\\ \Leftrightarrow \sin 2x\left( {2\cos x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin 2x = 0\\ 2\cos x + 1 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} 2x = k\pi \\ \cos x = \dfrac{{ - 1}}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{k\pi }}{2}\\ x = \pm \dfrac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in \mathbb{Z} } \right) \)
26 tháng 7 2019
\( 3)\sin x + \sin 2x + \sin 3x + \sin 4x = 0\\ \Leftrightarrow \left( {\sin x + \sin 4x} \right) + \left( {\sin 2x + \sin 3x} \right) = 0\\ \Leftrightarrow 2\sin \dfrac{{5x}}{2}.\cos \dfrac{{3x}}{2} + 2\sin \dfrac{{5x}}{2}.\cos \dfrac{x}{2} = 0\\ \Leftrightarrow \sin \dfrac{{5x}}{2}.\left( {\cos \dfrac{{3x}}{2} + \cos \dfrac{x}{2}} \right) = 0\\ \Leftrightarrow \sin \dfrac{{5x}}{2}.2\cos x.\cos \dfrac{x}{2} = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin \dfrac{{5x}}{2} = 0\\ 2\cos x = 0\\ \cos \dfrac{x}{2} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{{2k\pi }}{5}\\ x = \dfrac{\pi }{2} + k\pi \\ x = \pi + 2k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right) \)
21 tháng 7 2022
b: \(\Leftrightarrow2\cdot\cos2x\cdot\cos x+2\cdot\sin x\cdot\cos2x=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow2\cdot\cos2x\left(\sin x+\cos x\right)=\sqrt{2}\cdot\cos2x\)
\(\Leftrightarrow\sqrt{2}\cdot\cos2x\cdot\left[\sqrt{2}\cdot\sqrt{2}\cdot\sin\left(x+\dfrac{\Pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\cos2x=0\\\sin\left(x+\dfrac{\Pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\Pi}{2}+k\Pi\\x+\dfrac{\Pi}{4}=\dfrac{\Pi}{6}+k2\Pi\\x+\dfrac{\Pi}{4}=\dfrac{5}{6}\Pi+k2\Pi\end{matrix}\right.\)
\(\Leftrightarrow x\in\left\{\dfrac{\Pi}{4}+\dfrac{k\Pi}{2};\dfrac{-1}{12}\Pi+k2\Pi;\dfrac{7}{12}\Pi+k2\Pi\right\}\)
c: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 5 2020
\(A=\frac{sin3x-sinx+cos2x}{cosx-cos3x+sin2x}=\frac{2cos2x.sinx+cos2x}{2sin2x.sinx+sin2x}=\frac{cos2x\left(2sinx+1\right)}{sin2x\left(2sinx+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
15 tháng 8 2023
\(\Leftrightarrow sin2x\cdot sinx-cos2x\cdot sinx+sin2x\cdot cosx+sinx\cdot cos2x=cosx\left(sinx+cosx\right)\)
=>\(sin2x\left(sinx+cosx\right)=cosx\left(sinx+cosx\right)\)
=>\(\left(sinx+cosx\right)\cdot\left(sin2x-cosx\right)=0\)
=>\(cosx\cdot\left(2sinx-1\right)\cdot\sqrt{2}\cdot sin\left(x+\dfrac{pi}{4}\right)=0\)
=>\(\left[{}\begin{matrix}cosx=0\\2sinx-1=0\\sin\left(x+\dfrac{pi}{4}\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\sinx=\dfrac{1}{2}\\x+\dfrac{pi}{4}=kpi\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\x=-\dfrac{pi}{4}+kpi\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\x=-\dfrac{pi}{4}+kpi\\x=\dfrac{pi}{6}+k2pi\\x=\dfrac{5}{6}pi+k2pi\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{pi}{2}+kpi;-\dfrac{pi}{4}+kpi;\dfrac{pi}{6}+k2pi;\dfrac{5}{6}pi+k2pi\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 8 2020
c/
\(\Leftrightarrow cos3x-\sqrt{3}sin3x=\sqrt{3}cos2x-sin2x\)
\(\Leftrightarrow\frac{1}{2}cos3x-\frac{\sqrt{3}}{2}sin3x=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(3x+\frac{\pi}{3}\right)=cos\left(2x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{3}=2x+\frac{\pi}{6}+k2\pi\\3x+\frac{\pi}{3}=-2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=-\frac{\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 8 2020
b/
\(\Leftrightarrow cosx-\sqrt{3}sinx=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{5\pi}{6}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\sin3x+\sin x+\sin2x=0\)
\(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=0\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\Pi\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\Pi}{2}\\x=\dfrac{2}{3}\Pi+k2\Pi\\x=-\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)