Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
c/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)
\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)
\(\Leftrightarrow sin^3x-sin^2x=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=1\left(l\right)\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sin2x.sinx+cos2x.cosx}{sinx.cosx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}\)
\(\Leftrightarrow\frac{cos\left(2x-x\right)}{sinx.cosx}=\frac{sin^2x-cos^2x}{sinx.cosx}\)
\(\Leftrightarrow cosx=sin^2x-cos^2x\)
\(\Leftrightarrow cosx=1-2cos^2x\)
\(\Leftrightarrow2cos^2x+cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(l\right)\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y=\dfrac{xsinx}{tanx}+\dfrac{cosx}{tanx}=x.cosx+\dfrac{cos^2x}{sinx}=x.cosx+\dfrac{1}{sinx}-sinx\)
\(y'=cosx-x.sinx-\dfrac{cosx}{sin^2x}-cosx=-x.sinx-\dfrac{cosx}{sin^2x}\)
\(\Rightarrow y'+y.tan=-x.sinx-\dfrac{cosx}{sin^2x}+x.sinx+cosx\)
\(=cosx\left(1-\dfrac{1}{sin^2x}\right)=\dfrac{-cosx\left(1-sin^2x\right)}{sin^2x}=\dfrac{-cos^3x}{sin^2x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2.
$y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x$
$=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x$
Vì: $0\leq \sin ^22x\leq 1$
$\Rightarrow 1\geq 1-\frac{1}{2}\sin ^22x\geq \frac{1}{2}$
Vậy $y_{\max}=1; y_{\min}=\frac{1}{2}$
3.
$0\leq |\sin x|\leq 1$
$\Rightarrow 3\geq 3-2|\sin x|\geq 1$
Vậy $y_{\min}=1; y_{\max}=3$
![](https://rs.olm.vn/images/avt/0.png?1311)
a,
\(\cos^3x-\sin^3x=\cos x+\sin x\\ < =>\cos^3x-\cos x=\sin^3x-\sin x\\ < =>\cos x\left(\cos^2x-1\right)=\sin x\left(\sin^2x-1\right)\\ < =>\cos x.\left(-\sin^2x\right)=\sin x.\left(-\cos^2x\right)\\ < =>\dfrac{1}{cosx}=\dfrac{1}{sinx}\)
b,
\(2sinx+2\sqrt{3}cosx=\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}\\ < =>2sinx-\dfrac{1}{sinx}=\dfrac{\sqrt{3}}{cosx}-2\sqrt{3}cosx\\ < =>\dfrac{2sin^2x-1}{sinx}=\dfrac{\sqrt{3}.cosx.\left(1-2cos^2x\right)}{cosx}\\ < =>\dfrac{cos2x}{sinx}=\sqrt{3}.cos2x\\ < =>\dfrac{1}{sinx}=\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(y=sin2x+\left(\sqrt{3}+1\right)cos2x+sin^2x-cos^2x-1\)
\(=sin2x+\sqrt{3}cos2x-1=2sin\left(2x+\frac{\pi}{3}\right)-1\)
Do \(-1\le sin\left(2x+\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le1\)
b/ \(y=2sin^2x-2cos^2x-3sinx.cosx-1\)
\(=-2cos2x-\frac{3}{2}sin2x-1=-\frac{5}{2}\left(\frac{3}{5}sinx+\frac{4}{5}cosx\right)-1\)
\(=-\frac{5}{2}sin\left(x+a\right)-1\Rightarrow-\frac{7}{2}\le y\le\frac{3}{2}\)
c/ \(y=1-sin2x+2cos2x+\frac{3}{2}sin2x=\frac{1}{2}sin2x+2cos2x+1\)
\(=\frac{\sqrt{17}}{2}\left(\frac{1}{\sqrt{17}}sin2x+\frac{4}{\sqrt{17}}cos2x\right)+1=\frac{\sqrt{17}}{2}sin\left(2x+a\right)+1\)
\(\Rightarrow-\frac{\sqrt{17}}{2}+1\le y\le\frac{\sqrt{17}}{2}+1\)
Chọn D.
Bước đầu tiên sử dụng đạo hàm tổng, sau đó sử dụng (sin u)’, (cos u)’.
y' = (sin(cosx))’ + (cos(sinx))’ = cos(cosx).(cosx)’ – sin(sinx).(sinx)’
= -sinx.cos(cosx) – cosx.sin(sinx) = -(sinx.cos(cosx) + cosx.sin(sinx))
= -sin(x + cosx).