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HP
1 tháng 6 2021
1.
\(sinx-\sqrt{2}cos3x=\sqrt{3}cosx+\sqrt{2}sin3x\)
\(\Leftrightarrow sinx-\sqrt{3}cosx=\sqrt{2}cos3x+\sqrt{2}sin3x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{\sqrt{2}}cos3x+\dfrac{1}{\sqrt{2}}sin3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin\left(3x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=3x+\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{3}=\pi-3x-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7\pi}{24}-k\pi\\x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{7\pi}{24}-k\pi;x=-\dfrac{3}{4}x+\dfrac{13\pi}{48}+\dfrac{k\pi}{2}\)
HP
1 tháng 6 2021
2.
\(sinx-\sqrt{3}cosx=2sin5\text{}x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin5x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin5x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=5x+k2\pi\\x-\dfrac{\pi}{3}=\pi-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2021
a.
ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
Chia 2 vế cho cosx:
\(tanx+1=\dfrac{1}{cos^2x}\)
\(\Rightarrow tanx+1=1+tan^2x\)
\(\Rightarrow\left[{}\begin{matrix}tanx=0\\tanx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 9 2021
c.
\(\Leftrightarrow2sin2x+2sin^2x=1\)
\(\Leftrightarrow2sin2x=1-2sin^2x\)
\(\Leftrightarrow2sin2x=cos2x\)
\(\Rightarrow tan2x=\dfrac{1}{2}\)
\(\Rightarrow2x=arctan\left(\dfrac{1}{2}\right)+k\pi\)
\(\Rightarrow x=\dfrac{1}{2}arctan\left(\dfrac{1}{2}\right)+\dfrac{k\pi}{2}\)
D
1
22 tháng 10 2017
\(sinx+cosx\cdot sin2x+\sqrt{3}cos3x=2.\left(cos4x+sin^3x\right)\)
\(\Leftrightarrow sinx+cosx\cdot sin2x+\sqrt{3}cos3x=2cos4x+2sin^3x\)
\(\Leftrightarrow sinx-2sin^3x+cosx.sin2x+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow sinx.\left(1-2sin^2x\right)+cosx.sin2x+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow sinx.cos2x+cosx.sin2x+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow sin.\left(x+2x\right)+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow sin3x+\sqrt{3}cos3x=2cos4x\)
\(\Leftrightarrow\dfrac{1}{2}sin3x+\dfrac{\sqrt{3}}{2}cos3x=cos4x\)
\(\Leftrightarrow cos\dfrac{\pi}{3}.sin3x+sin\dfrac{\pi}{3}.cos3x=cos4x\)
\(\Leftrightarrow sin.\left(3x+\dfrac{\pi}{3}\right)=sin\left(\dfrac{\pi}{x}-4x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}-4x+k2\pi\\3x+\dfrac{\pi}{2}=\pi-\dfrac{\pi}{2}+4x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{42}+\dfrac{k2\pi}{7}\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
d/
\(\Leftrightarrow2\left(sinx-cosx\right)\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(sinx-cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\2\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow2+2sinx.cosx=\sqrt{3}cos2x\)
\(\Leftrightarrow2+sin2x=\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=-1\)
\(\Leftrightarrow2x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{12}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
31 tháng 7 2020
c/
\(\Leftrightarrow sinx-sin^2x=cosx-cos^2x\)
\(\Leftrightarrow sinx-cosx-\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow sinx-cosx-\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(1-sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\1-sinx-cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\\1-\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
14 tháng 8 2017
a, \(sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2cos^2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-2\cdot\left[1+cos2\cdot\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right]=0\)
\(\Leftrightarrow sin\dfrac{x}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x+1-1-cos\left(\dfrac{\pi}{2}-x\right)=0\)
\(\Leftrightarrow sin\dfrac{s}{2}\cdot sinx-cos\dfrac{x}{2}\cdot sin^2x-sinx=0\)
\(\Leftrightarrow sinx\cdot\left(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\text{ (1) }\\sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx=0\Leftrightarrow x=k\pi\left(k\in Z\right)\)
(2) : \(sin\dfrac{x}{2}-sinx\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-cos\dfrac{x}{2}\cdot2sin\dfrac{x}{2}\cdot cos\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot cos^2\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}\cdot\left(1-sin^2\dfrac{x}{2}\right)-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}-2sin\dfrac{x}{2}+2sin^3\dfrac{x}{2}-1=0\)
\(\Leftrightarrow2sin^3\dfrac{x}{2}-sin\dfrac{x}{2}-1=0\)
\(\Leftrightarrow sin\dfrac{x}{2}=1\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\pi+k4\pi\left(k\in Z\right)\)
14 tháng 8 2017
b, \(tanx-3cotx=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cos}{sinx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sinx-cosx}=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\)
\(\Leftrightarrow sin^2x-3cos^2x=4\cdot\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx-\sqrt{3}\cdot cosx\right)\cdot\left(sinx+\sqrt{3}\cdot cosx\right)=4\left(sinx+\sqrt{3}\cdot cosx\right)\cdot sinx\cdot cosx\)
\(\Leftrightarrow\left(sinx+\sqrt{3}\cdot cosx\right)\cdot\left[\left(sinx-\sqrt{3}\cdot cosx\right)-4sinx\cdot cosx\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}\cdot cosx=0\text{ (1) }\\sinx-\sqrt{3}\cdot cosx-4sinx\cdot cosx=0\text{ (2) }\end{matrix}\right.\)
(1) : \(sinx+\sqrt{3}\cdot cosx=0\)
\(\Leftrightarrow\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\)
\(\Leftrightarrow cos\dfrac{\pi}{3}\cdot sinx+sin\dfrac{\pi}{3}\cdot cosx=0\)
\(\Leftrightarrow sin\cdot\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow x+\dfrac{\pi}{3}=k\pi\Leftrightarrow x=\dfrac{-\pi}{3}+k\pi\left(k\in Z\right)\)
(2) : \(sinx-\sqrt{3}cosx-4sinx\cdot cosx=0\)
\(\Leftrightarrow sinx-\sqrt{3}cos=2sin2x\)
\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cos2=sin2x\)
\(\Leftrightarrow cos\dfrac{\pi}{3}-sinx-sin\dfrac{\pi}{3}\cdot cosx=sin2x\)
\(\Leftrightarrow sin\cdot\left(x-\dfrac{\pi}{3}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{3}=2x+k2\pi\\x-\dfrac{\pi}{3}=\pi-2x+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\left(k\in Z\right)\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 9 2019
a/ ĐKXĐ: \(\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k2\pi\\x\ne-\frac{\pi}{6}+k2\pi\\x\ne\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(1+sinx-2sin^2x\right)\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(cos2x+sinx\right)\)
\(\Leftrightarrow\sqrt{3}sinx-cosx=sin2x+\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}sin2x+\frac{\sqrt{3}}{2}cos2x\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=sin\left(2x+\frac{\pi}{6}\right)\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(cosx+\sqrt{3}sinx\ne0\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)\ne0\Rightarrow...\)
Đặt \(cosx+\sqrt{3}sinx=2sin\left(x+\frac{\pi}{6}\right)=a\) với \(-2\le a\le2\):
\(a=\frac{3}{a}+1\Leftrightarrow a^2-a-3=0\)
\(\Rightarrow\left[{}\begin{matrix}a=\frac{1+\sqrt{13}}{2}>2\left(l\right)\\a=\frac{1-\sqrt{13}}{2}\end{matrix}\right.\)
\(\Rightarrow2sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{2}\)
\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{13}}{4}=sin\alpha\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=\alpha+k2\pi\\x+\frac{\pi}{6}=\pi-\alpha+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 6 2019
Đặt \(\left|sinx-cosx\right|=a\) (\(0\le a\le\sqrt{2}\))
\(\Rightarrow1-2sinx.cosx=a^2\Rightarrow1-sin2x=a^2\Rightarrow sin2x=1-a^2\)
Phương trình trở thành:
\(a+4\left(1-a^2\right)=1\Leftrightarrow-4a^2+a+3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{4}< 9\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left|sinx-cosx\right|=1\Leftrightarrow\left|\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\right|=1\)
\(\Leftrightarrow\left|sin\left(x-\frac{\pi}{4}\right)\right|=\frac{\sqrt{2}}{2}\Rightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\\sin\left(x-\frac{\pi}{4}\right)=\frac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Rightarrow...\)
NM
1
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 4 2021
ĐKXĐ: ...
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx=\dfrac{3}{2}\left(1+tan^2x\right)-\sqrt{3}tanx\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=\dfrac{3}{2}\left(tanx-\dfrac{\sqrt{3}}{3}\right)^2+1\)
\(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)\le1\\\dfrac{3}{2}\left(tanx-\dfrac{\sqrt{3}}{3}\right)^2+1\ge1\end{matrix}\right.\)
Đẳng thức xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=1\\tanx=\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{\pi}{6}+k2\pi\)
3 tháng 4 2021
Hình như có nhầm lẫn từ dòng 1 xuống dòng 2 thì phải. Em bấm máy tính ra nghiệm pi/6 mà.
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2020
d/
ĐKXĐ: \(cosx\ne0\)
\(\Leftrightarrow\frac{sin\left(3x-x\right)}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sin2x}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{2sinx.cosx}{cos^2x}=2\sqrt{3}\)
\(\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\)
\(\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 7 2020
c/
ĐKXĐ: \(sin2x\ne0\)
\(\Leftrightarrow\frac{\frac{sinx}{cosx}-sinx}{sin^3x}=\frac{1}{cosx}\)
\(\Leftrightarrow sinx-sinx.cosx=sin^3x\)
\(\Leftrightarrow1-cosx=sin^2x\)
\(\Leftrightarrow1-cosx=1-cos^2x\)
\(\Leftrightarrow cos^2x-cosx=0\Rightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left(\sqrt{3}+2\right)sinx+cosx=2sin3x+2sinx\)
\(\Leftrightarrow\sqrt{3}sinx+cosx=2sin3x\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=sin3x\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=sin3x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=x+\dfrac{\pi}{6}+k2\pi\\3x=\dfrac{5\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow x=...\)