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NH
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 8 2020
\(1+sinx-cos2x=0\)
\(\Leftrightarrow1+sinx-\left(1-2sin^2x\right)=0\)
\(\Leftrightarrow sinx\left(1+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(sin3x-sinx+cos2x=0\)
\(\Leftrightarrow2cos2x.sinx+cos2x=0\)
\(\Leftrightarrow cos2x\left(2sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
QA
1
28 tháng 7 2020
\(sin3x\left(cosx-sin3x\right)+cos3x\left(sinx-cos3x\right)=0\\ \Leftrightarrow sin3x\cdot cosx+cos3x\cdot sinx=sin^23x+cos^23x\\ \Leftrightarrow sin4x=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{2}\)
21 tháng 7 2022
\(\Leftrightarrow\sin3x+\sin x+\sin2x=0\)
\(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=0\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\Pi\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\Pi}{2}\\x=\dfrac{2}{3}\Pi+k2\Pi\\x=-\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
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
11 tháng 8 2020
5.
\(sin^2x+sin^22x=1\)
\(\Leftrightarrow4sin^2x.cos^2x-\left(1-sin^2x\right)=0\)
\(\Leftrightarrow4sin^2x.cos^2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4sin^2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=\frac{1}{2}\\sinx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=...\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 8 2020
1.
\(\Leftrightarrow\frac{1}{2}\left(cos8x+cos6x\right)=\frac{1}{2}\left(cos8x+cos2x\right)\)
\(\Leftrightarrow cos6x=cos2x\)
\(\Rightarrow\left[{}\begin{matrix}6x=2x+k2\pi\\6x=-2x+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{k\pi}{4}\end{matrix}\right.\) \(\Rightarrow x=\frac{k\pi}{4}\)
2.
\(sin3x-4sinx.cos2x=0\)
\(\Leftrightarrow sin3x-2\left(sin3x-sinx\right)=0\)
\(\Leftrightarrow-sin3x+2sinx=0\)
\(\Leftrightarrow4sin^3x-3sinx+2sinx=0\)
\(\Leftrightarrow sinx\left(4sin^2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\frac{1}{2}\\sinx=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow x=...\)
JE
2
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 7 2020
b/
\(\Leftrightarrow sin3x-sinx-sin3x=1\)
\(\Leftrightarrow sinx=-1\)
\(\Rightarrow x=-\frac{\pi}{2}+k2\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 7 2020
a/ \(\Leftrightarrow sin3x+sinx-sin2x=0\)
\(\Leftrightarrow2sin2x.cosx-sin2x=0\)
\(\Leftrightarrow sin2x\left(2cosx-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=0\\2cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=k\pi\\x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2021
\(sin5x+sinx+sin3x=0\)
\(\Leftrightarrow2sin3x.cos2x+sin3x=0\)
\(\Leftrightarrow sin3x\left(2cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=k\pi\\2x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{3}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 8 2020
ĐKXĐ:
\(sin3x-sinx\ne0\)
\(\Leftrightarrow sin3x\ne sinx\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne\pi-x+n2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{\pi}{4}+\frac{n\pi}{2}\end{matrix}\right.\)
\(sin3x=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=x+k2\pi\\3x=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)
\(\sin3x-\sin x=0\)
\(\Leftrightarrow\sin3x=\sin x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=x+k2\pi\\3x=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k2\pi\\4x=\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)