Tìm min, max
a) \(y=\frac{8}{3-cos^2x}\)
b) \(\frac{1}{\sqrt{2-sin^23x}}\)
c) \(y=\sqrt{3}\left(cos^4x-sin^4x\right)+sin2x+1\)
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) \(4sinx.cosx.cos2x+cos4x=\sqrt{2}\)
\(\text{a) }cos^2x+sin2x-1=0\\ \Leftrightarrow2sinx\cdot cosx-sin^2x=0\\ \Leftrightarrow sinx\left(2cosx-sinx\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=2cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=0\\tanx=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=a\pi\\x=arctan\left(2\right)+b\pi\end{matrix}\right.\)
\(\text{b) }\sqrt{3}sin2x+cos^4x-sin^4x=\sqrt{2}\\ \Leftrightarrow\sqrt{3}sin2x+\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=\sqrt{2}\\ \Leftrightarrow\frac{\sqrt{3}}{2}\cdot sin2x+\frac{1}{2}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sin2x+sin\frac{\pi}{6}\cdot cos2x=\frac{\sqrt{2}}{2}\\ \Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=sin\frac{\pi}{4}\\ \\ \Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=\frac{\pi}{4}+a2\pi\\2x+\frac{\pi}{6}=\frac{3\pi}{4}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{24}+a\pi\\x=\frac{7\pi}{24}+b\pi\end{matrix}\right.\)
\(c\text{) }cos^2x-sin^2x=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\\ \Leftrightarrow cos^2x-sin^2x=\sqrt{2}\left(sinx\cdot\frac{\sqrt{2}}{2}+cosx\cdot\frac{\sqrt{2}}{2}\right)\\ \Leftrightarrow\left(cosx-sinx\right)\left(sinx+cosx\right)=sinx+cosx\\ \Leftrightarrow\left[{}\begin{matrix}cosx-sinx=1\\sinx=-cosx\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cos^2x+\left(cosx-1\right)^2=1\\tanx=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\tanx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+a\pi\\x=b2\pi\\x=\frac{3\pi}{4}=c\pi\end{matrix}\right.\)
\(d\text{) }4\left(sin^4x+cos^4x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow4\left(1-2sin^2x\cdot cos^2x\right)+\sqrt{3}sin4x=2\\ \Leftrightarrow-8sin^2x\cdot cos^2x+\sqrt{3}sin4x=-2\\ \Leftrightarrow-2sin^22x+\sqrt{3}sin4x=-2\\ \Leftrightarrow cos4x-1+\sqrt{3}sin4x=-2\\ \Leftrightarrow\frac{1}{2}cos4x+\frac{\sqrt{3}}{2}sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\frac{\pi}{6}\cdot cos4x+cos\frac{\pi}{6}\cdot sin4x=-\frac{1}{2}\\ \Leftrightarrow sin\left(4x+\frac{\pi}{6}\right)=sin\frac{-\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}4x+\frac{\pi}{6}=\frac{-\pi}{6}+a2\pi\\4x+\frac{\pi}{6}=\frac{7\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-\pi}{12}+\frac{a\pi}{2}\\x=\frac{\pi}{4}+\frac{b\pi}{2}\end{matrix}\right.\)
\(e\text{) }4sinx\cdot cosx\cdot cos2x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x+cos4x=\sqrt{2}\\ \Leftrightarrow sin4x\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}cos4x=1\\ \Leftrightarrow sin4x\cdot cos\frac{\pi}{4}+cos4x\cdot sin\frac{\pi}{4}=1\\ \Leftrightarrow sin\left(4x+\frac{\pi}{4}\right)=1=sin\frac{\pi}{2}\\ \Leftrightarrow4x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{\pi}{16}+\frac{k\pi}{2}\)
giải pt
a) \(sin^2x+2sin^22x+sin^23x-2=0\)
b) \(2cosx.cos\left(x+\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)
c) \(5\left(1+cosx\right)=2+sin^4x-cos^4x\)
d) \(1+cot2x=\frac{1-cos2x}{sin^22x}\)
a/
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x+\frac{1}{2}-\frac{1}{2}cos6x-2\left(1-sin^22x\right)=0\)
\(\Leftrightarrow1-\frac{1}{2}\left(cos6x+cos2x\right)-2cos^22x=0\)
\(\Leftrightarrow1-cos4x.cos2x-2cos^22x=0\)
\(\Leftrightarrow2cos^22x-1+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x+cos4x.cos2x=0\)
\(\Leftrightarrow cos4x\left(cos2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos2x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+k\pi\\2x=\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{k\pi}{4}\\x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+cos\left(\frac{\pi}{3}\right)+\sqrt{3}sin2x=1\)
\(\Leftrightarrow cos2x.cos\left(\frac{\pi}{3}\right)-sin2x.sin\left(\frac{\pi}{3}\right)+\frac{1}{2}+\sqrt{3}sin2x=1\)
\(\Leftrightarrow\frac{1}{2}cos2x+\frac{\sqrt{3}}{2}sin2x=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{3}+k2\pi\\2x-\frac{\pi}{3}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+k\pi\\x=k\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow5+5cosx=2+\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)\)
\(\Leftrightarrow3+5cosx=sin^2x-cos^2x\)
\(\Leftrightarrow3+5cosx=1-cos^2x-cos^2x\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-2\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
giải các pt
a) \(4sin^3x+3\sqrt{2}sin2x=8sinx\)
b) \(7cosx=4cos^3x+4sin2x\)
c) \(tanx+cotx=5-\frac{3}{sin^22x}\)
d) \(5\left(1+cosx\right)=2+sin^4x-cos^4x\)
e) \(2\left(cos^2x+cos^22x+cos^23x\right)=3\left(1+cosx.cos4x\right)\)
a/
\(\Leftrightarrow4sin^3x+6\sqrt{2}sinx.cosx-8sinx=0\)
\(\Leftrightarrow2sinx\left(2sin^2x+3\sqrt{2}cosx-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\2sin^2x+3\sqrt{2}cosx-4=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2\left(1-cos^2x\right)+3\sqrt{2}cosx-4=0\)
\(\Leftrightarrow-2cos^2x+3\sqrt{2}cosx-2=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=\sqrt{2}>1\left(l\right)\\cosx=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{4}+k2\pi\)
b/
\(\Leftrightarrow4cos^3x+8sinx.cosx-7cosx=0\)
\(\Leftrightarrow cosx\left(4cos^2x+8sinx-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\4cos^2x+8sinx-7=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow4\left(1-sin^2x\right)+8sinx-7=0\)
\(\Leftrightarrow-4sin^2x+8sinx-3=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=\frac{3}{2}\left(l\right)\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
c/
ĐKXĐ; ...
\(\Leftrightarrow\frac{sinx}{cosx}+\frac{cosx}{sinx}-5+\frac{3}{sin^22x}=0\)
\(\Leftrightarrow\frac{sin^2x+cos^2x}{sinx.cosx}-5+\frac{3}{sin^22x}=0\)
\(\Leftrightarrow\frac{3}{sin^22x}+\frac{2}{sin2x}-5=0\)
Đặt \(\frac{1}{sin2x}=t\Rightarrow3t^2+2t-5=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{5}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{1}{sin2x}=1\\\frac{1}{sin2x}=-\frac{5}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=-\frac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(-\frac{3}{5}\right)+k\pi\end{matrix}\right.\)
Giải các phương trình :
1) \(\frac{\sin^4x+\cos^4x}{\sin2x}=\frac{1}{2}\left(\tan x+\cot2x\right)\)
2) \(\frac{1}{\sin x}+\frac{1}{\sin\left(x-\frac{3\pi}{2}\right)}=4\sin\left(\frac{7\pi}{4}-x\right)\)
3)\(2\left(\cos^42x-\sin^42x\right)+\cos8x-\cos4x=0\)
4)\(\frac{\sin^4x+\cos^4x}{5\sin2x}=\frac{1}{2}\cot2x-\frac{1}{8\sin2x}\)
5)\(\sin^4x+\cos^4x-3\sin2x+\frac{5}{2}\sin^22x=0\)
\(\sqrt{\sin^4x+4\cos^2x}+\sqrt{\cos^4x+4\sin^2x}\)
=\(\sqrt{\left(1-cos^2x\right)^2+4\cos^2x}+\sqrt{\left(1-sin^2x\right)^2+4\sin^2x}\)
=\(\sqrt{\cos^4x-2\cos^2x+1+4\cos^2x}+\sqrt{\sin^4x-2\sin^2x+1+4\sin^2x}\)
=\(\sqrt{\cos^4x+2\cos^2x+1}+\sqrt{\sin^4x+2\sin^2x+1}\)
=\(\sqrt{\left(cos^2x+1\right)^2}+\sqrt{\left(sin^2x+1\right)^2}\)
=\(cos^2x+1+sin^2x+1=3\)
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 \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
Tìm Max, Min của hàm số:
1) \(y=\dfrac{x+1+\sqrt{x-1}}{x+1+2\sqrt{x-1}}\)
2) \(y=\sin^{2016}x+\cos^{2016}x\)
3) \(y=2\cos x-\dfrac{4}{3}\cos^3x\) trên \(\left[0;\dfrac{\pi}{2}\right]\)
4) \(y=\sin2x-\sqrt{2}x+1,x\in\left[0;\dfrac{\pi}{2}\right]\)
5) \(y=\dfrac{4-cos^2x}{\sqrt{sin^4x+1}},x\in\left[-\dfrac{\pi}{3};\dfrac{\pi}{3}\right]\)
Giải phương trình:
1.\(cos^3x.cos3x+sin^3x.sin3x=\frac{\sqrt{2}}{4}\)
2.\(cos^34x=cos^3x.cos3x+sin^3x.sin3x\)
3.\(cos^2x-4sin^2\left(\frac{x}{2}-\frac{\pi}{4}\right)+2=0\)
4.\(sin^4x+sin^4\left(x+\frac{\pi}{4}\right)=\frac{1}{4}\)
5.\(sin^6x+cos^6x=\frac{5}{6}\left(sin^4x+cos^4x\right)\)
6.\(sin^6x+cos^6x+\frac{1}{2}sinx.cosx=0\)
7.\(\frac{1}{2}\left(sin^4x+cos^4x\right)=sin^2x.cos^2x+sinx.cosx\)
8.\(sin^6x+cos^6x-3cos8x+2=0\)
9.\(sin^4x+cos^4x-2\left(sin^6\frac{x}{2}+cos^6\frac{x}{2}\right)+1=0\)
5.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\frac{5}{6}\left[\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\right]\)
\(\Leftrightarrow1-3sin^2x.cos^2x=\frac{5}{6}\left(1-2sin^2x.cos^2x\right)\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x=\frac{5}{6}\left(1-\frac{1}{2}sin^22x\right)\)
\(\Leftrightarrow\frac{1}{3}sin^22x=\frac{1}{6}\)
\(\Leftrightarrow sin^22x=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\frac{\sqrt{2}}{2}\\sin2x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\\x=\frac{5\pi}{8}+k\pi\end{matrix}\right.\)
6.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+\frac{1}{2}sinx.cosx=0\)
\(\Leftrightarrow1-\frac{3}{4}sin^22x+\frac{1}{4}sin2x=0\)
\(\Leftrightarrow-3sin^22x+sin2x+4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=-1\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow2x=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{4}+k\pi\)
1.
\(\Rightarrow4cos^3x.cos3x+4sin^3x.sin3x=\sqrt{2}\)
\(\Leftrightarrow\left(3cosx+cos3x\right)cos3x+\left(3sinx-sin3x\right)sin3x=\sqrt{2}\)
\(\Leftrightarrow3\left(cos3x.cosx+sin3x.sinx\right)+cos^23x-sin^23x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+cos6x=\sqrt{2}\)
\(\Leftrightarrow3cos2x+4cos^32x-3cos2x=\sqrt{2}\)
\(\Leftrightarrow4cos^32x=\sqrt{2}\)
\(\Leftrightarrow cos2x=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{4}+k2\pi\\2x=-\frac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)
Tìm giá trị max, min của các hàm số sau:
1, y= 2 - \(\sin\left(\dfrac{3\pi}{2}+x\right)\cos\left(\dfrac{\pi}{2}+x\right)\)
2, y= \(\sqrt{5-2\sin^2x.\cos^2x}\)
1, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)
\(y=2-\left(-cosx\right).\left(-sinx\right)\)
y = 2 - sinx.cosx
y = \(2-\dfrac{1}{2}sin2x\)
Max = 2 + \(\dfrac{1}{2}\) = 2,5
Min = \(2-\dfrac{1}{2}\) = 1,5
2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)
Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)
Max = \(\sqrt{5}\)