tìm gtln gtnn của hàm số
a,\(\frac{4}{1+2sin^2x}\)
b, \(\left(4sin-3cos\right)^2-4\left(4sinx-3cosx\right)+1\)
Giải các pt sau
a, \(\dfrac{1}{sinx}+\dfrac{1}{cosx}=4sin\left(x+\dfrac{\pi}{4}\right)\)
b, \(2sin\left(2x-\dfrac{\pi}{6}\right)+4sinx+1=0\)
c, \(cos2x+\sqrt{3}sinx+\sqrt{3}sin2x-cosx=2\)
d, \(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+cos^2\left(x-\dfrac{3\pi}{4}\right)\)
Tìm giá trị lớn nhất và giá trị nhỏ nhất của các hàm số sau:
1,\(y=5-3cosx\)
2,\(y=3cos^2x-2cosx+2\)
3,\(y=cos^2x+2cos2x\)
4,\(y=\sqrt{5-2sin^2x.cos^2x}\)
5,\(y=cos2x-cos\left(2x-\dfrac{\pi}{3}\right)\)
6,\(y=\sqrt{3}sinx-cosx-2\)
7,\(y=2cos^2x-sin2x+5\)
8,\(y=2sin^2x-sin2x+10\)
9,\(y=sin^6x+cos^6x\)
Tìm GTLN; GTNN của các hàm số:
\(a,y=2sin^2x-cos2x\)
\(b,y=3\sqrt{1+sinx}-1\) trên đoạn \(\left[0;\dfrac{\pi}{3}\right]\)
a, \(y=2sin^2x-cos2x=1-2cos2x\)
Vì \(cos2x\in\left[-1;1\right]\Rightarrow y=2sin^2x-cos2x\in\left[-1;3\right]\)
\(\Rightarrow\left\{{}\begin{matrix}y_{min}=-1\\y_{max}=3\end{matrix}\right.\)
Tìm GTLN; GTNN của các hàm số
\(a,y=3-4sin^2xcos^2x\)
\(b,y=\dfrac{-2}{3sinx-5}\) trên đoạn \(\left[0;\dfrac{\pi}{2}\right]\)
a, \(y=3-4sin^2x.cos^2x=3-sin^22x\)
Đặt \(sin2x=t\left(t\in\left[-1;1\right]\right)\).
\(\Rightarrow y=f\left(t\right)=3-t^2\)
\(\Rightarrow y_{min}=minf\left(t\right)=2\)
\(y_{max}=maxf\left(t\right)=3\)
b, \(y=f\left(t\right)=\dfrac{-2}{3t-5}\left(t\in\left[0;1\right]\right)\)
\(\Rightarrow y_{min}=minf\left(t\right)=\dfrac{2}{5}\)
\(y_{max}=maxf\left(t\right)=1\)
Tìm GTNN, GTLN của hàm số y= 2sin2x + 3cosx -1 trên đoạn \(\left[\dfrac{-\pi}{3};\dfrac{2\pi}{3}\right]\)
Tìm GTLN và GTNN của hàm số:
y=\(\sqrt{5-2sin^2xcos^2x}\)
y= sinx trên \(\left[\frac{\pi}{6};\frac{3\pi}{4}\right]\)
giải phương trình:
a, \(tanx.sin^2x-2sin^2x=3\left(cos2x+sinxcosx\right)\)
b, \(5sinx-2=3\left(1-sinx\right)tan^2x\)
c,\(\frac{cos2x+3cot2x+4sinx}{cot2x-cos2x}=2\)
d, \(\frac{4sin^2x+6sin^2x-3cos2x-9}{cosx}=0\)
a)\(4sin^3xcos3x+4cos^3xsin3x+3\sqrt{3}cos4x=3\)
b)\(2sin^2x\left(4sin^4x-1\right)=cos2x\left(7cos^22x+3cos2x-4\right)\)
2 câu này giải như nào ạ
a
\(\Leftrightarrow\left(3sinx-sin3x\right)cos3x+\left(3cosx+cos3x\right)sin3x+3\sqrt{3}cos4x=3\)
\(\Leftrightarrow\left(sinx.cos3x+sin3x.cosx\right)+\sqrt{3}cos4x=1\)
\(\Leftrightarrow sin4x+\sqrt{3}cos4x=1\)
Tới đây thôi, mình lười ghi rồi =))
b
\(\Leftrightarrow\left(1-cos2x\right)\left(2sin^2x-1\right)\left(2sin^2+1\right)=cos2x\left(7cos^22x+3cos2x-4\right)\)
\(\Leftrightarrow\left(1-cos2x\right)\left(-cos2x\right)\left(2-cos2x\right)=cos2x\left(7cos^22x+3cos2x+4\right)\)
\(\Leftrightarrow-cos^22x+3cos2x-2=7cos^22x+3cos2x+4\)
\(\Leftrightarrow4cos^22x+3=0\)
=> pt vô nghiệm
giải các pt
a) \(cos^2\left(\frac{\pi}{3}+x\right)+4cos\left(\frac{\pi}{6}-x\right)=4\)
b) \(5cos\left(2x+\frac{\pi}{3}\right)=4sin\left(\frac{5\pi}{6}-x\right)-9\)
c) \(2sin^2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
d) \(2sin^2x+\sqrt{3}sin2x+4=4\left(\sqrt{3}sinx+cosx\right)\)
a/
Đặt \(x+\frac{\pi}{3}=a\Rightarrow x=a-\frac{\pi}{3}\)
Pt trở thành:
\(cos^2a+4cos\left(\frac{\pi}{6}-a+\frac{\pi}{3}\right)=4\)
\(\Leftrightarrow cos^2a+4cos\left(\frac{\pi}{2}-a\right)-4=0\)
\(\Leftrightarrow cos^2a+4sina-4=0\)
\(\Leftrightarrow1-sin^2a+4sina-4=0\)
\(\Leftrightarrow-sin^2a+4sina-3=0\)
\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sin\left(x+\frac{\pi}{3}\right)=1\)
\(\Rightarrow x+\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{6}+k2\pi\)
b/
Đặt \(x+\frac{\pi}{6}=a\Rightarrow x=a-\frac{\pi}{6}\)
Pt trở thành:
\(5cos2a=4sin\left(\frac{5\pi}{6}-a+\frac{\pi}{6}\right)-9\)
\(\Leftrightarrow5cos2x=4sin\left(\pi-a\right)-9\)
\(\Leftrightarrow5\left(1-2sin^2a\right)=4sina-9\)
\(\Leftrightarrow10sin^2a+4sina-14=0\)
\(\Rightarrow\left[{}\begin{matrix}sina=1\\sina=-\frac{7}{5}< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sin\left(x+\frac{\pi}{6}\right)=1\)
\(\Rightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=\frac{\pi}{3}+k2\pi\)
c/
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x+2\sqrt{3}sinx+2cosx=2\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x+2\left(\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow cos2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow1-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}=0\)
\(\Leftrightarrow-2sin^2\left(x+\frac{\pi}{6}\right)+2sin\left(x+\frac{\pi}{6}\right)+\frac{1}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{6}\right)=\frac{1+\sqrt{2}}{2}\left(l\right)\\sin\left(x+\frac{\pi}{6}\right)=\frac{1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{6}=arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\\x+\frac{\pi}{6}=\pi-arcsin\left(\frac{1-\sqrt{2}}{2}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)