sin4x + cos4x = 4\(\sqrt{2}\)sin(x+\(\frac{\pi}{4}\))
1) \(\frac{1}{\cos x}+\frac{1}{\sin2x}=\frac{2}{\sin4x}\)
2) \(\cos3x\cdot\tan5x=\sin7x\)
3) \(\tan5x\cdot\tan2x=1\)
4) \(4\cos x-2\cos2x-\cos4x=1\)
5) \(\sin\left(2x+\frac{5\pi}{2}\right)-2\cos\left(x-\frac{7\pi}{2}\right)=1+2\sin x\)
6) \(\sin^22x-\cos^28x=\sin\left(\frac{17\pi}{2}+10x\right)\)
7) \(8\cos x=\frac{\sqrt{3}}{\sin x}+\frac{1}{\cos x}\)
1.
DKXĐ: \(sin4x\ne0\)
\(\Leftrightarrow\frac{4sinx.cos2x}{sin4x}+\frac{2cos2x}{sin4x}=\frac{2}{sin4x}\)
\(\Leftrightarrow2sinx.cos2x+cos2x=1\)
\(\Leftrightarrow2sinx\left(1-2sin^2x\right)+1-2sin^2x=1\)
\(\Leftrightarrow sinx\left(1-2sin^2x-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\-2sin^2x-sinx+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}sinx=-1\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
2.
ĐKXĐ: ...
\(\Leftrightarrow\frac{cos3x.sin5x}{cos5x}=sin7x\)
\(\Leftrightarrow cos3x.sin5x=sin7x.cos5x\)
\(\Leftrightarrow sin8x+sin2x=sin12x+sin2x\)
\(\Leftrightarrow sin8x=sin12x\)
\(\Leftrightarrow\left[{}\begin{matrix}12x=8x+k2\pi\\12x=\pi-8x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{2}\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)
Ở nghiệm đầu tiên loại các giá trị k lẻ do đó nghiệm của pt là:
\(\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{20}+\frac{k\pi}{10}\end{matrix}\right.\)
3.
ĐKXĐ: ...
\(\Leftrightarrow tan5x=\frac{1}{tan2x}\)
\(\Leftrightarrow tan5x=cot2x\)
\(\Leftrightarrow tan5x=tan\left(\frac{\pi}{2}-2x\right)\)
\(\Leftrightarrow5x=\frac{\pi}{2}-2x+k\pi\)
\(\Leftrightarrow x=\frac{\pi}{14}+\frac{k\pi}{7}\)
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 pt
a) \(cos^2x+sin2x-1=0\)
b) \(\sqrt{3}sin2x+\:cos^4x-sin^4x=\sqrt{2}\)
c) \(\:cos^2x-sin^2x=\sqrt{2}.sin\left(x+\frac{\pi}{4}\right)\)
d) \(4\left(sin^4x+cos^4x\right)+\sqrt{3}.sin4x=2\)
e) \(4sinx.cosx.cos2x+cos4x=\sqrt{2}\)
\(\text{a) }cos^2x+sin2x-1=0\\ \Leftrightarrow2sinx\cdot cosx-sin^2x=0\\ \Leftrightarrow sinx\left(2cosx-sinx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=2cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=a\pi\\x=arctan\left(2\right)+b\pi\end{matrix}\right.\)
\(\text{b) }\sqrt{3}sin2x+cos^4x-sin^4x=\sqrt{2}\\ \Leftrightarrow\sqrt{3}sin2x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\sqrt{2}\\ \Leftrightarrow\frac{\sqrt{3}}{2}\cdot sin2x+\frac{1}{2}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{4}\\ \\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{4}+a2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+a\pi\\x=\frac{7\pi}{24}+b\pi\end{matrix}\right.\)
\(c\text{) }cos^2x-sin^2x=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\\ \Leftrightarrow cos^2x-sin^2x=\sqrt{2}\left(sinx\cdot\frac{\sqrt{2}}{2}+cosx\cdot\frac{\sqrt{2}}{2}\right)\\ \Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)=sinx+cosx\\ \Leftrightarrow\left[{}\begin{matrix}cosx-sinx=1\\sinx=-cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos^2x+\left(cosx-1\right)^2=1\\tanx=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\tanx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+a\pi\\x=b2\pi\\x=\frac{3\pi}{4}=c\pi\end{matrix}\right.\)
\(d\text{) }4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow4\left(1-2sin^2x\cdot cos^2x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow-8sin^2x\cdot cos^2x+\sqrt{3}sin4x=-2\\ \Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\\ \Leftrightarrow cos4x-1+\sqrt{3}sin4x=-2\\ \Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\frac{\pi}{6}\cdot cos4x+cos\frac{\pi}{6}\cdot sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=sin\frac{-\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=\frac{-\pi}{6}+a2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{12}+\frac{a\pi}{2}\\x=\frac{\pi}{4}+\frac{b\pi}{2}\end{matrix}\right.\)
\(e\text{) }4sinx\cdot cosx\cdot cos2x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cos4x=1\\ \Leftrightarrow sin4x\cdot cos\frac{\pi}{4}+cos4x\cdot sin\frac{\pi}{4}=1\\ \Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{16}+\frac{k\pi}{2}\)
Tìm nghiệm của pt:
1) \(2cos2x+\sqrt{2}cos\frac{\pi}{4}=0\) thuộc khoảng (0;2π)
2) \(sin4x-cos4x+\sqrt{2}cos\left(4x-\frac{\pi}{4}\right)=\sqrt{6}\) thuộc khoảng (-π;5π)
1.
\(\Leftrightarrow2cos2x+\sqrt{2}.\frac{\sqrt{2}}{2}=0\)
\(\Leftrightarrow cos2x=-\frac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
\(\Rightarrow x=\left\{\frac{\pi}{3};\frac{4\pi}{3};\frac{2\pi}{3};\frac{5\pi}{3}\right\}\)
2.
\(\Leftrightarrow sin4x-cos4x+sin4x+cos4x=\sqrt{6}\)
\(\Leftrightarrow2sin4x=\sqrt{6}\)
\(\Leftrightarrow sin4x=\frac{\sqrt{6}}{2}>1\)
Pt vô nghiệm
\(cosx-2cos3x=1+\sqrt{3}sinx\)
\(sinx+sinx\left(x+\dfrac{\pi}{3}\right)+sin4x=sin\left(2x-\dfrac{\pi}{3}\right)\)
\(\left(1-\dfrac{1}{2sinx}\right)cos^22x=2sinx-3+\dfrac{1}{sinx}\)
( sinx -2cosx)cos2x + sinx = (cos4x - 1)cosx +\(\dfrac{cos2x}{2sinx}\)
\(\left(\dfrac{cos4x+sin2x}{cos3x+sin3x}\right)^2=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\)
Trong khoảng \(\left(0;\frac{\pi}{2}\right)\) thì pt: \(\sin^24x+3\sin4x.\cos4x-4\cos^24x=0\) có bn nghiệm?
\(\Leftrightarrow\left(sin4x-cos4x\right)\left(sin4x+4cos4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x-cos4x=0\\sin4x=-4cos4x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(4x-\frac{\pi}{4}\right)=0\\tan4x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{4}=k\pi\\4x=arctan\left(-4\right)+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{16}+\frac{k\pi}{4}\\x=\frac{1}{4}arctan\left(-4\right)+\frac{k\pi}{4}\end{matrix}\right.\)
Pt có 4 nghiệm trên khoảng đã cho
Giải các phương trình sau
1) \(\sin^6x+\cos^6x=\cos4x\)
2) \(\sin^2\left(2x+\frac{5\pi}{12}\right)+\sin^2\left(x+\frac{\pi}{12}\right)=1\)
3) \(\sin x.\cos4x-\sin^22x=4\sin^2\left(\frac{\pi}{4}-\frac{x}{2}\right)-\frac{7}{2}\)
bài 1 bung công thức sin^6(x) + cos^6(x) là 5/8 + 3/8cos4x = cos4x chuyển vế giải
bài 2 dùng công thức hạ bậc sau đó dùng công thức cộng là ra
giải phương trình
\(\sin x\sqrt{1+2\sin x}=\cos2x\)
\(\sin\left(\frac{5x}{2}-\frac{\pi}{4}\right)-\cos\left(\frac{x}{2}-\frac{\pi}{4}\right)=\sqrt{2}\cos\frac{3x}{2}\)
\(3\sqrt{\tan x+1}\left(\sin x+2\cos x\right)=5\left(\sin x+3\cos x\right)\)
\(\sqrt{2}\left(\sin x+\sqrt{3}\cos x\right)=\sqrt{3}\cos2x-\sin2x\)
\(\sin2x\sin4x+2\left(3\sin x-4\sin^2x+1\right)=0\)
a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
a)sin^4\(\frac{x}{3}\) +cos^4\(\frac{x}{3}\)=\(\frac{5}{8}\)
b)4(sin^4x+cos^4x)+\(\sqrt{3}\)sin4x=2
c)cos^4x+sin^6x=cos2x
d)cos^6x+sin^6x=cos4x
2cos^2x+2cos^2x+4cos^3(2x)-3cos2x=5
a/
\(\Leftrightarrow\left(sin^2\frac{x}{3}+cos^2\frac{x}{3}\right)^2-2sin^2\frac{x}{3}.cos^2\frac{x}{3}=\frac{5}{8}\)
\(\Leftrightarrow1-\frac{1}{2}sin^2\frac{2x}{3}=\frac{5}{8}\)
\(\Leftrightarrow1-\frac{1}{4}\left(1-cos\frac{4x}{3}\right)=\frac{5}{8}\)
\(\Leftrightarrow cos\frac{4x}{3}=-\frac{1}{2}\)
\(\Leftrightarrow\frac{4x}{3}=\pm\frac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\frac{\pi}{2}+\frac{k3\pi}{2}\)
b/
\(\Leftrightarrow4\left(sin^2x+cos^2x\right)^2-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)
\(\Leftrightarrow4-8sin^2x.cos^2x+\sqrt{3}sin4x=2\)
\(\Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x=-1\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x+\frac{1}{2}cos4x=-\frac{1}{2}\)
\(\Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=-\frac{\pi}{6}+k2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+\frac{k\pi}{2}\\x=\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\)
c/
\(\left(\frac{1+cos2x}{2}\right)^2+\left(\frac{1-cos2x}{2}\right)^3=cos2x\)
\(\Leftrightarrow-cos^32x+5cos^22x-7cos2x+3=0\)
\(\Leftrightarrow\left(3-cos2x\right)\left(cos2x-1\right)^2=0\)
\(\Leftrightarrow cos2x=1\)
\(\Leftrightarrow x=k\pi\)
d/
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos4x\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=cos4x\)
\(\Leftrightarrow1-\frac{3}{8}\left(1-cos4x\right)=cos4x\)
\(\Leftrightarrow cos4x=1\)
\(\Leftrightarrow x=\frac{k\pi}{2}\)