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28 tháng 7 2020
\(sin3x\left(cosx-sin3x\right)+cos3x\left(sinx-cos3x\right)=0\\ \Leftrightarrow sin3x\cdot cosx+cos3x\cdot sinx=sin^23x+cos^23x\\ \Leftrightarrow sin4x=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{8}+\frac{k\pi}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 8 2020
\(1+sinx-cos2x=0\)
\(\Leftrightarrow1+sinx-\left(1-2sin^2x\right)=0\)
\(\Leftrightarrow sinx\left(1+2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(sin3x-sinx+cos2x=0\)
\(\Leftrightarrow2cos2x.sinx+cos2x=0\)
\(\Leftrightarrow cos2x\left(2sinx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 9 2021
\(sin5x+sinx+sin3x=0\)
\(\Leftrightarrow2sin3x.cos2x+sin3x=0\)
\(\Leftrightarrow sin3x\left(2cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=k\pi\\2x=\pm\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{3}\\x=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
21 tháng 7 2022
\(\Leftrightarrow\sin3x+\sin x+\sin2x=0\)
\(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=0\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k\Pi\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\Pi}{2}\\x=\dfrac{2}{3}\Pi+k2\Pi\\x=-\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
CM
6 tháng 6 2019
Chọn A
Ta có sin3x+ cos2x- sinx= 0 ⇔ cos2x(2sinx+1)=0. Lưu ý trong khoảng (0;π), sinx > 0
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
1 tháng 8 2021
ĐKXĐ: ...
\(sin3x-cos3x+sinx+cosx=\dfrac{sin3x-cos3x+sinx+cosx}{\left(sin3x+cosx\right)\left(cos3x-sinx\right)}\)
\(\Rightarrow\left[{}\begin{matrix}sin3x-cos3x+sinx+cosx=0\left(1\right)\\\left(sin3x+cosx\right)\left(cos3x-sinx\right)=1\left(2\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow3sinx-4sin^3x-4cos^3x+3cosx+sinx+cosx=0\)
\(\Leftrightarrow sinx+cosx+sin^3x+cos^3x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)\left(1-sinx.cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2-sinx.cosx\right)=0\)
\(\Leftrightarrow sinx+cosx=0\) (loại)
(2) \(\Leftrightarrow sin3x.cos3x-sinx.cosx-sin3x.sinx+cos3x.cosx=1\)
\(\Leftrightarrow\dfrac{1}{2}sin6x-\dfrac{1}{2}sin2x+cos4x=1\)
\(\Leftrightarrow\dfrac{1}{2}\left(3sin2x-4sin^32x\right)-\dfrac{1}{2}sin2x+1-2sin^22x=1\)
\(\Leftrightarrow sin2x-2sin^32x-2sin^22x=0\)
\(\Leftrightarrow-sin2x\left(2sin^22x+2sin2x-1\right)=0\)
\(\Leftrightarrow...\)
\(sin3x=sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=x+k2\pi\\3x=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)
\(\sin3x-\sin x=0\)
\(\Leftrightarrow\sin3x=\sin x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=x+k2\pi\\3x=\pi-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=k2\pi\\4x=\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)