giải các pt
a) \(sin2x-2sinx=0\)
b) \(cosx.cos4x-cosx.cos2x=0\)
c) \(cos11x.cos3x=cos17x.cos9x\)
d) \(sin18x.cos13x=sin9x.cos4x\)
giải các pt
a) \(tanx-\frac{\sqrt{2}}{cosx}=1\)
b) \(\frac{2sinx-1}{cos4x}+\frac{2sinx-1}{sin4x-1}=0\)
c) \(sin\left(x+\frac{\pi}{4}\right)-cos\left(x-\frac{\pi}{4}\right)=1\)
d) \(\frac{sin2x-2cos2x-5}{2sin2x-cos2x-6}=0\)
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: ...
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)
\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)
\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)
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/
\(\Leftrightarrow sin2x-2cos2x-5=2sin2x-cos2x-6\)
\(\Leftrightarrow sin2x+cos2x=1\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
c/
Hình như câu này đề sai
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)-\sqrt{2}cos\left(x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sinx+cosx-\left(sinx+cosx\right)=\sqrt{2}\)
\(\Leftrightarrow0=\sqrt{2}\)
Pt vô nghiệm
d/ Hình như câu này đề cũng sai
\(\Leftrightarrow sin2x-2cos2x-5=0\)
\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x-\frac{2}{\sqrt{5}}cos2x=\sqrt{5}\)
\(\Leftrightarrow sin\left(2x-a\right)=\sqrt{5}\) (với \(sina=\frac{2}{\sqrt{5}};cosa=\frac{1}{\sqrt{5}}\))
Pt vô nghiệm do \(\sqrt{5}>1\)
Giải các phương trình sau:
a, \(\sqrt{2}\) sin \(\left(2x+\frac{\pi}{4}\right)\)=3sinx+cosx+2
b, 1+sinx+cosx+sin2x+cos2x=0
c, (2cosx-1)(2sinx+cosx)=sin2x-sinx
d, cos3x+cos2x-cosx-1=0
a.
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(sinx+cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=-1\\2cosx-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\\cosx=\frac{3}{2}\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
\(\Leftrightarrow1+sinx+cosx+2sinx.cosx+2cos^2x-1=0\)
\(\Leftrightarrow sinx\left(2cosx+1\right)+cosx\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c.
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giúp mình với
Giải pt sau:
\(sin2x\left(2cosx+1\right)-cos2x-2sinx-cosx+1=0\)
Giải phương trình sin2x + 2sinx – 3=0
giải các pt (nhân tử chung)
a) \(sin4x=2cos2x.cosx\)
b) \(\left(cosx+sin2x\right).sin2x=0\)
c) \(cosx+cos2x+cos3x+cos4x=0\)
d) \(sin3x-sinx+sin2x=0\)
\(sin4x-2cos2x.cosx=0\)
\(\Leftrightarrow2sin2x.cos2x-2cos2x.cosx=0\)
\(\Leftrightarrow cos2x\left(sin2x-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\sin2x-cosx=0\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow sin2x=cosx=sin\left(\frac{\pi}{2}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\left(cosx+sin2x\right).sin2x=0\)
\(\Leftrightarrow\left(cosx+2sinx.cosx\right).2sinx.cosx=0\)
\(\Leftrightarrow\left(1+2sinx\right)sinx.cos^2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1+2sinx=0\\sinx.cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\2x=k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\end{matrix}\right.\)
\(cosx+cos3x+cos2x+cos4x=0\)
\(\Leftrightarrow2cos2x.cosx+2cos3x.cosx=0\)
\(\Leftrightarrow cosx\left(cos2x+cos3x\right)=0\)
\(\Leftrightarrow2cosx.cos\frac{5x}{2}.cos\frac{x}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\\cos\frac{x}{2}=0\\cos\frac{5x}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{5x}{2}=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\frac{\pi}{5}+\frac{k2\pi}{5}\end{matrix}\right.\)
1/Giải phương trình:
a/ sin5x.cos3x = sin7x.cos5x
b/ \(\sqrt{2}\left(sinx-2cosx\right)=2-sin2x\)
2/ Phương trình \(2sinx\left(1+cos2x\right)=1+2cosx\) là phương trình hệ quả của phương trình:
A. cos2x = 0
B. 2cosx - 1 = 0
C. sin2x + 1 = 0
D. sin2x - 1 = 0
a/ \(\Leftrightarrow sin8x+sin2x=sin12x+sin2x\)
\(\Leftrightarrow sin12x=sin8x\)
\(\Rightarrow\left[{}\begin{matrix}12x=8x+k2\pi\\12x=\pi-8x+k2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)
b/ \(\sqrt{2}sinx-2\sqrt{2}cosx-2+2sinx.cosx=0\)
\(\Leftrightarrow\sqrt{2}sinx\left(\sqrt{2}cosx+1\right)-2\left(\sqrt{2}cosx+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}sinx-2\right)\left(\sqrt{2}cosx+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=\sqrt{2}>1\left(l\right)\\cosx=-\frac{\sqrt{2}}{2}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{3\pi}{4}+k2\pi\)
c/ Không là hệ quả của pt nào, chắc bạn ghi nhầm đề bài
giải pt 2sinx + cos3x +sin2x=1+sin4x
\(\Leftrightarrow2sinx+cos3x+sin2x-sin4x-1=0\)
\(\Leftrightarrow2sinx-1+cos3x-2cos3x.sinx=0\)
\(\Leftrightarrow2sinx-1-cos3x\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(1-cos3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\cos3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k2\pi}{3}\end{matrix}\right.\)
Giải PT
a) sin2 x + 2sinx - 3 = 0
b) 2cos x + cos 2x = 0
c) tanx + cotx + 2 = 0
d) sinx + cos2x + 1 = 0
e) tan x + cot 2x = 0
a) TH1: sinx = 1
--> x = pi/2 + k2pi (k nguyên)
TH2: sinx = -3 (loại)
b) 2cosx + cos2x = 0
<=> 2cosx + 2cos^2(x) - 1 = 0
TH1: cosx = (-1 + sqrt(3))/2
TH2: cosx = (-1 - sqrt(3))/2 (loại)
c) ĐKXĐ: x # kpi
Pt <=> tanx + 1/tanx + 2 = 0
--> tanx = -1
--> x = -pi/4 + kpi (k nguyên)
Giải phương trình
1,sin3x+cos2x=1+2sinx*cos2x
2,cos5x+cos2x+2sin3x*sin2x=0