\(\left(2sin5x-1\right).\left(2cos2x-1\right)=2sinx\)
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NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
c/
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sinx=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow sin2x-2cos2x-5=2sin2x-cos2x-6\)
\(\Leftrightarrow sin2x+cos2x=1\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
a/ ĐKXĐ:...
\(\Leftrightarrow\frac{sinx}{cosx}-\frac{\sqrt{2}}{cosx}=1\)
\(\Leftrightarrow sinx-\sqrt{2}=cosx\)
\(\Leftrightarrow sinx-cosx=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{3\pi}{4}+k2\pi\)
b/
ĐKXĐ: ...
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)
\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)
\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 7 2021
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k2\pi\\x\ne-\dfrac{\pi}{6}+k2\pi\\x\ne\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\dfrac{cosx-2sinx.cosx}{1-2sin^2x+sinx}=\sqrt{3}\)
\(\Leftrightarrow\dfrac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)
\(\Rightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=-x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\left(loại\right)\\x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
KB
17 tháng 7 2021
ĐKXĐ : \(sinx\ne1;-\dfrac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+2k\pi\\x\ne\dfrac{-\pi}{6}+2k\pi;\dfrac{7\pi}{6}+2k\pi\end{matrix}\right.\)
\(\Leftrightarrow x\ne\dfrac{-\pi}{6}+\dfrac{2}{3}k\pi\)( k thuộc Z )
P/t đã cho \(\Leftrightarrow\dfrac{cosx-sin2x}{1-2sin^2x+sinx}=\sqrt{3}\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(cos2x+sinx\right)\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x+\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+2k\pi\\2x+\dfrac{\pi}{6}=-x-\dfrac{\pi}{3}+2k\pi\end{matrix}\right.\) ( k thuộc Z )
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+2k\pi\\x=\dfrac{-\pi}{6}+\dfrac{2}{3}k\pi\left(L\right)\end{matrix}\right.\)
Vậy ...
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 10 2020
4.
\(\Leftrightarrow2sinx.cosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+2sin^2x+3sinx-2=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+\left(2sinx-1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\sinx+cosx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 10 2020
2.
ĐKXĐ: ...
\(\Leftrightarrow cot\left(\frac{\pi}{4}-x\right)=-\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow\frac{\pi}{4}-x=-\frac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\frac{7\pi}{12}+k\pi\)
3.
\(\Leftrightarrow cos\frac{x}{4}sinx+sin\frac{x}{4}.cosx-3\left(sin^2x+cos^2x\right)+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{x}{4}\right)=-cosx\)
\(\Leftrightarrow sin\frac{5x}{4}=sin\left(x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{5x}{4}=x-\frac{\pi}{2}+k2\pi\\\frac{5x}{4}=\frac{3\pi}{2}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
18 tháng 6 2017
\(pt\Leftrightarrow\dfrac{\left(1-2\sin x\right)\cos x}{1-\sin^2x}=\sqrt{3}\Leftrightarrow\dfrac{1-2\sin x}{\sqrt{3}\cos x}=1\)
\(\Leftrightarrow1-2\sin x=\sqrt{3}\cos x\Leftrightarrow\sqrt{3}\cos x+2\sin x=1\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{\sqrt{7}}\cos x+\dfrac{2}{\sqrt{7}}\sin x=1\)
\(\Leftrightarrow\cos a\cdot\cos x+\sin a\cdot\sin x=1\) với \(a=\sin^{-1}\dfrac{2}{\sqrt{7}}=\cos^{-1}\dfrac{\sqrt{3}}{\sqrt{7}}\)
\(\Leftrightarrow\cos\left(a-x\right)=1\Leftrightarrow a-x=k2\pi\)
\(\Leftrightarrow x=a-k2\pi\Leftrightarrow x=a+m2\pi\left(m\in Z\right)\)
P.s: lâu lâu pick thử bài lượng phát, nguy cơ đúng 80% nhé :)
BC
1
15 tháng 8 2023
=>\(2\cdot cos2x\cdot sin2x+2cos^22x-sin2x-cos2x-1=0\)
=>\(2cos2x\cdot sin2x+2\cdot cos^22x-1=sin2x+cos2x\)
=>\(sin4x+cos4x=sin2x+cos2x\)
=>\(sin\left(4x+\dfrac{pi}{4}\right)=sin\left(2x+\dfrac{pi}{4}\right)\)
=>4x+pi/4=2x+pi/4+k2pi hoặc 4x+pi/4=pi-2x-pi/4+k2pi
=>2x=k2pi hoặc 6x=1/2pi+k2pi
=>x=kpi hoặc x=1/12pi+kpi/3
MA
1
HP
17 tháng 8 2021
\(\left(2cos2x-1\right)\left(sin2x+cos2x\right)=1\)
\(\Leftrightarrow2sin2x.cos2x+2cos^22x-sin2x-cos2x-1=0\)
\(\Leftrightarrow sin4x+cos4x-sin2x-cos2x=0\)
\(\Leftrightarrow2cos3x.sinx-2sin3x.sinx=0\)
\(\Leftrightarrow2sinx\left(cos3x-sin3x\right)=0\)
\(\Leftrightarrow2\sqrt{2}sinx.cos\left(3x+\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos\left(3x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\3x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\end{matrix}\right.\)
NA
Giải pt:
\(\left(2sinx-1\right)^2-\left(2sinx-1\right)\left(sinx-\frac{3}{2}\right)=0\)
Giúp với ạ !
1
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 9 2020
\(\Leftrightarrow\left(2sinx-1\right)\left(2sinx-1-sinx+\frac{3}{2}\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sinx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
Với \(cosx=0\) ko phải nghiệm
Với \(cosx\ne0\)
\(\Rightarrow\left(2sin5x-1\right)\left(2cos2x.cosx-cosx\right)=2sinx.cosx\)
\(\Leftrightarrow\left(2sin5x-1\right)\left(cos3x+cosx-cosx\right)=sin2x\)
\(\Leftrightarrow cos3x\left(2sin5x-1\right)=sin2x\)
\(\Leftrightarrow2sin5x.cos3x-cos3x=sin2x\)
\(\Leftrightarrow sin8x+sin2x-cos3x=sin2x\)
\(\Leftrightarrow sin8x=cos3x=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow...\)
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