Giải các phương trình:
\(a,sinx-cosx=1\)
\(b,cos3x.cos2x=cos5x\)
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N
11 tháng 9 2016
a)pt\(\Leftrightarrow cosx\left(cosx+1\right)+sinx.sin^2x=0\)
\(\Leftrightarrow cosx\left(cosx+1\right)+sinx\left(1-cos^2x\right)=0\)
\(\Leftrightarrow\left(cosx+1\right)\left(cosx+sinx-sinx.cosx\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}cosx=1\Leftrightarrow x=\pi+k2\pi\\cosx+sinx-sinx.cosx=0\left(\cdot\right)\end{array}\right.\)
Xét pt(*):
Đặt \(t=cosx+sinx,t\in\left[-\sqrt{2};\sqrt{2}\right]\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)
(*) trở thành:\(t^2-2t-1=0\Leftrightarrow\left[\begin{array}{nghiempt}t=1-\sqrt{2}\\t=1+\sqrt{2}\left(L\right)\end{array}\right.\)
+)\(t=1-\sqrt{2}\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\\ \Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{\pi}{4}+arcsin\left(\frac{-2+\sqrt{2}}{2}\right)+k2\pi\\x=-\frac{5\pi}{4}-arcsin\left(\frac{-2+\sqrt{2}}{2}\right)+k2\pi\end{cases}\left(k\in Z\right)}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
4.
ĐKXĐ: \(2cos^2x+sinx-1\ne0\)
\(\Leftrightarrow-2sin^2x+sinx+1\ne0\Rightarrow\left\{{}\begin{matrix}sinx\ne1\\sinx\ne-\frac{1}{2}\end{matrix}\right.\)
Khi đó pt tương đương:
\(\Leftrightarrow\frac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)
\(\Leftrightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)
\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)
\(\Leftrightarrow\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx=\frac{\sqrt{3}}{2}cos2x+\frac{1}{2}sin2x\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{3}\right)=cos\left(2x-\frac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{6}=x+\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{6}=-x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\left(loại\right)\\x=-\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 8 2020
3.
\(\Leftrightarrow cos7x+\sqrt{3}sin7x=sin5x+\sqrt{3}cos5x\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin7x+\frac{1}{2}cos7x=\frac{1}{2}sin5x+\frac{\sqrt{3}}{2}cos5x\)
\(\Leftrightarrow sin\left(7x+\frac{\pi}{6}\right)=sin\left(5x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+\frac{\pi}{6}=5x+\frac{\pi}{3}+k2\pi\\7x+\frac{\pi}{6}=\frac{2\pi}{3}-5x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{\pi}{24}+\frac{k\pi}{6}\end{matrix}\right.\)
20 tháng 8 2021
a) Đặt \(sinx+cosx=t\left(\left|t\right|\le\sqrt{2}\right)\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)
=> pt có dạng: \(t=\sqrt{2}\left(t^2-1\right)\Leftrightarrow\sqrt{2}t^2-t-\sqrt{2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=\frac{-\sqrt{2}}{2}\\t=\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}sinx+cosx=\frac{-\sqrt{2}}{2}\\sinx+cosx=\sqrt{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}sin\left(x+\frac{\pi}{4}\right)=\frac{-1}{2}\\sin\left(x+\frac{\pi}{4}\right)=1\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x+\frac{\pi}{4}=\frac{-\pi}{6}+2k\pi\\x+\frac{\pi}{4}=\frac{7\pi}{6}+2k\pi\\x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{-5\pi}{12}+2k\pi\\x=\frac{11\pi}{12}+2k\pi\\x=\frac{\pi}{4}+2k\pi\end{cases}}\left(k\inℤ\right)}\)
HP
13 tháng 8 2021
1.
a, Phương trình có nghiệm khi:
\(\left(m+2\right)^2+m^2\ge4\)
\(\Leftrightarrow m^2+4m+4+m^2\ge4\)
\(\Leftrightarrow2m^2+4m\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}m\ge0\\m\le-2\end{matrix}\right.\)
b, Phương trình có nghiệm khi:
\(m^2+\left(m-1\right)^2\ge\left(2m+1\right)^2\)
\(\Leftrightarrow2m^2+6m\le0\)
\(\Leftrightarrow-3\le m\le0\)
HP
13 tháng 8 2021
2.
a, Phương trình vô nghiệm khi:
\(\left(2m-1\right)^2+\left(m-1\right)^2< \left(m-3\right)^2\)
\(\Leftrightarrow4m^2-4m+1+m^2-2m+1< m^2-6m+9\)
\(\Leftrightarrow4m^2-7< 0\)
\(\Leftrightarrow-\dfrac{\sqrt{7}}{2}< m< \dfrac{\sqrt{7}}{2}\)
b, \(2sinx+cosx=m\left(sinx-2cosx+3\right)\)
\(\Leftrightarrow\left(m-2\right)sinx-\left(2m+1\right)cosx=-3m\)
Phương trình vô nghiệm khi:
\(\left(m-2\right)^2+\left(2m+1\right)^2< 9m^2\)
\(\Leftrightarrow m^2-4m+4+4m^2+4m+1< 9m^2\)
\(\Leftrightarrow m^2-1>0\)
\(\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)
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 \(x=-\dfrac{\pi}{12}-\dfrac{k\pi}{2};x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
PT
1
không thỏa mãn
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) \(\Leftrightarrow sin3x.cos3x-sinx.cosx-sin3x.sinx+cos3x.cosx=1\)
\(\Leftrightarrow\dfrac{1}{2}sin6x-\dfrac{1}{2}sin2x+cos4x=1\)
\(\Leftrightarrow\dfrac{1}{2}\left(3sin2x-4sin^32x\right)-\dfrac{1}{2}sin2x+1-2sin^22x=1\)
\(\Leftrightarrow sin2x-2sin^32x-2sin^22x=0\)
\(\Leftrightarrow-sin2x\left(2sin^22x+2sin2x-1\right)=0\)
\(\Leftrightarrow...\)
a, \(sinx-cosx=1\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
b, \(cos3x.cos2x=cos5x\)
\(\Leftrightarrow\dfrac{1}{2}cos5x+\dfrac{1}{2}cosx=cos5x\)
\(\Leftrightarrow cosx=cos5x\)
\(\Leftrightarrow5x=\pm x+k2\pi\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{k\pi}{3}\end{matrix}\right.\)