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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/ \(\Leftrightarrow sin3x+sinx-sin2x=0\)
\(\Leftrightarrow2sin2x.cosx-sin2x=0\)
\(\Leftrightarrow sin2x\left(2cosx-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=0\\2cosx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=k\pi\\x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
VL
2
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2020
1.
\(\Leftrightarrow cos3x+sin3x-2sin3x.cos3x=0\)
\(\Leftrightarrow cos3x+sin3x-\left(2sin3x.cos3x+1\right)+1=0\)
\(\Leftrightarrow cos3x+sin3x-\left(sin3x+cos3x\right)^2+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x+cos3x=\frac{\sqrt{5}+1}{2}\\sin3x+cos3x=\frac{1-\sqrt{5}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(3x+\frac{\pi}{4}\right)=\frac{\sqrt{10}+\sqrt{2}}{4}>1\left(l\right)\\sin\left(3x+\frac{\pi}{4}\right)=\frac{\sqrt{2}-\sqrt{10}}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\frac{\pi}{4}=arcsin\left(\frac{\sqrt{2}-\sqrt{10}}{4}\right)+k2\pi\\3x+\frac{\pi}{4}=\pi-arcsin\left(\frac{\sqrt{2}-\sqrt{10}}{4}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 10 2020
2.
\(\Leftrightarrow sinx-\left(1+cosx\right)+sin2x=-2\)
\(\Leftrightarrow sinx-cosx+1+sin2x=0\)
\(\Leftrightarrow sinx-cosx-\left(1-2sinx.cosx\right)+2=0\)
\(\Leftrightarrow sinx-cosx-\left(sinx-cosx\right)^2+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=-1\\sinx-cosx=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}>1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
c/
\(\Leftrightarrow sinx+sin3x+sin2x=cosx+cos3x+cos2x\)
\(\Leftrightarrow2sin2x.cosx+sin2x=2cos2x.cosx+cos2x\)
\(\Leftrightarrow sin2x\left(2cosx+1\right)=cos2x\left(2cosx+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx+1=0\\sin2x=cos2x=sin\left(\frac{\pi}{2}-2x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\2x=\frac{\pi}{2}-2x+k2\pi\\2x=2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{2\pi}{3}+k2\pi\\x=\frac{\pi}{8}+\frac{k\pi}{2}\\\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2020
b/
\(\Leftrightarrow sin2x+sin6x-\left(cos5x+cosx\right)=0\)
\(\Leftrightarrow2sin4x.cos2x-2cos3x.cos2x=0\)
\(\Leftrightarrow cos2x\left(sin4x-cos3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin4x=cos3x=sin\left(\frac{\pi}{2}-3x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\4x=\frac{\pi}{2}-3x+k2\pi\\4x=3x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{14}+\frac{k2\pi}{7}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 8 2020
ĐKXĐ: ...
\(5\left(\frac{sinx+2sin2x.sinx+sin3x+cos3x}{1+2sin2x}\right)=cos2x+3\)
\(\Leftrightarrow5\left(\frac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}\right)=cos2x+3\)
\(\Leftrightarrow5\left(\frac{sinx+sin3x+cosx}{1+2sin2x}\right)=cos2x+3\)
\(\Leftrightarrow5\left(\frac{2sin2x.cosx+cosx}{1+2sin2x}\right)=cos2x+3\)
\(\Leftrightarrow5cosx=cos2x+3\)
\(\Leftrightarrow5cosx=2cos^2x+2\)
\(\Leftrightarrow2cos^2x-5cosx+2=0\Leftrightarrow cosx=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 7 2020
\(sin3x-sinx+sin2x=0\)
\(\Leftrightarrow2cos2x.sinx+2sinx.cosx=0\)
\(\Leftrightarrow sinx\left(cos2x+cosx\right)=0\)
\(\Leftrightarrow2sinx.cos\frac{3x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=0\\cos\frac{x}{2}=0\\cos\frac{3x}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{3x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\pi+k2\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+\frac{k2\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 7 2020
\(cosx+cos3x+cos2x+cos4x=0\)
\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)
\(\Leftrightarrow cosx\left(cos2x+cos3x\right)=0\)
\(\Leftrightarrow2cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{x}{2}=0\\cos\frac{5x}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
c/
\(\Leftrightarrow\sqrt{3}sin3x-cos3x=sin2x-\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin3x-\frac{1}{2}cos3x=\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{6}\right)=sin\left(2x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=2x-\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{6}=\pi-2x+\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 7 2020
e/
\(\Leftrightarrow\frac{1}{2}sin8x-\frac{\sqrt{3}}{2}cos8x=\frac{\sqrt{3}}{2}sin6x+\frac{1}{2}cos6x\)
\(\Leftrightarrow sin\left(8x-\frac{\pi}{3}\right)=sin\left(6x+\frac{\pi}{6}\right)\)
\(\Rightarrow\left[{}\begin{matrix}8x-\frac{\pi}{3}=6x+\frac{\pi}{6}+k2\pi\\8x-\frac{\pi}{3}=\pi-6x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{\pi}{28}+\frac{k\pi}{7}\end{matrix}\right.\)
5 tháng 9 2020
ngại viết quá hihi, mà hơi ngáo tí cái dạng này lm rồi mà cứ quên
bài trước mk bình luận bạn đọc chưa nhỉ
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: \(\Leftrightarrow2\cdot\sin2x\cdot\cos x+\sin2x=2\cdot\cos2x\cdot\cos x+\cos2x\)
\(\Leftrightarrow\sin2x\left(2\cos x+1\right)=\cos2x\left(2\cos x+1\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\sin2x=\cos2x=\sin\left(\dfrac{\Pi}{2}-2x\right)\\\cos x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{8}+\dfrac{k\Pi}{4}\\\\x=-\dfrac{2}{3}\Pi+k2\Pi\\x=\dfrac{2}{3}\Pi+k2\Pi\end{matrix}\right.\)
Mik chưa học lớp 11 nên ko trả lời đc sorry nha !! mik mới học lớp 6 thui
uk ko sao