limx-0\(\frac{x.sin3x}{1-cosx}\)
\(1.\left(sinx+cosx\right)^3+sinxcosx-1=0\)
\(2.\left(sinx+cosx\right)^4-3sin2x-1=0\)
\(3.sin^3x+cos^3x+2\left(sinx+cosx\right)-3sin2x=0\)
\(4.\left(sinx-cosx\right)^3=1+sinxcosx\)
5.\(sinx+cosx+2+tanx+cotx+\frac{1}{sinx}+\frac{1}{cosx}=0\)
1.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
Pt trở thành:
\(t^3+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow2t^3+t^2-3=0\)
\(\Leftrightarrow\left(t-1\right)\left(2t^2+3t+3\right)=0\)
\(\Leftrightarrow t=1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sin2x=2sinx.cosx=t^2-1\end{matrix}\right.\)
Pt trở thành:
\(t^4-3\left(t^2-1\right)-1=0\)
\(\Leftrightarrow t^4-3t^2+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t^2=1\\t^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}1+sin2x=1\\1+sin2x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
3.
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-sinx.cosx\right)+2\left(sinx+cosx\right)-6sinx.cosx=0\)
Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
Pt trở thành:
\(t\left(1-\frac{t^2-1}{2}\right)+2t-3\left(t^2-1\right)=0\)
\(\Leftrightarrow-t^3-6t^2+7t+6=0\)
Nghiệm của pt bậc 3 này rất xấu, chắc bạn ghi ko đúng đề bài
Cho \(0< x< 90^0\) CMR:
a) \(\sqrt{\frac{1+cosx}{1-cosx}}-\sqrt{\frac{1-cosx}{1+cosx}}=2cotx\)
b) \(\frac{1}{tanx+1}+\frac{1}{cotx+1}=1\)
c) \(sin^6x+cos^6x+sin^4x.cos^4x+5sin^2x.cos^2x=2\)
d) \(sin^21^0+sin^22^0+sin^23^0+...+sin^289^0=?\)
ai giúp mị với
a) \(\sqrt{\frac{1+\cos x}{1-\cos x}}-\sqrt{\frac{1-\cos x}{1+\cos x}}=\frac{\sqrt{\left(1+\cos x\right)^2}-\sqrt{\left(1-\cos x\right)^2}}{\sqrt{\left(1-\cos x\right)\left(1+\cos x\right)}}\)
\(=\frac{1+\cos x-1+\cos x}{\sqrt{1-\cos^2x}}=\frac{2\cos x}{\sqrt{\sin^2x}}=\frac{2\cos x}{\sin x}=2\cot x\)
b) \(\frac{1}{\tan x+1}+\frac{1}{\cot x+1}=\frac{\tan x+1+\cot x+1}{\left(\tan x+1\right)\left(\cot x+1\right)}\)
\(=\frac{\tan x+\cot x+2}{\tan x+\cot x+\tan x.\cot x+1}=\frac{\tan x+\cot x+2}{\tan x+\cot x+2}=1\)
c) (ko bt có sai đề ko, làm mãi ko ra)
d) \(\sin^21^0+\sin^22^0+\sin^23^0+...+\sin^289^0\)
\(=\left(\sin^21^0+\sin^289^0\right)+\left(\sin^22^0+\sin^288^0\right)+...+\sin^245^0\)
\(=\left[\left(\sin^21^0-\cos^289^0\right)+\left(\sin^289^0+\cos^289^0\right)\right]+\)
\(\left[\left(\sin^22^0-\cos^288^0\right)+\left(\sin^288^0+\cos^288^0\right)\right]+...+\sin^245^0\)
\(=\left(0+1\right)+\left(0+1\right)+...+\frac{\sqrt{2}}{2}=\frac{44+\sqrt{2}}{2}\)
giải các pt
a) \(\sqrt{3}sinx+cosx=\frac{1}{cosx}\)
b) \(-\frac{1}{2}tan^2x+\frac{2}{cosx}-\frac{5}{2}=0\)
a/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{3}\frac{sinx}{cosx}+1=\frac{1}{cos^2x}\)
\(\Leftrightarrow\sqrt{3}tanx+1=1+tan^2x\)
\(\Leftrightarrow tanx\left(tanx-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=0\\tanx=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
b/ ĐKXĐ: ...
\(\Leftrightarrow tan^2x+1-\frac{4}{cosx}+4=0\)
\(\Leftrightarrow\frac{1}{cos^2x}-\frac{4}{cosx}+4=0\)
\(\Leftrightarrow\left(\frac{1}{cosx}-2\right)^2=0\)
\(\Leftrightarrow\frac{1}{cosx}=2\)
\(\Rightarrow cosx=\frac{1}{2}\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
tìm GTLN,GTNN của hàm số sau:
a, \(y=\frac{1}{sinx}+\frac{1}{cosx},x\in\left(0,\frac{\pi}{2}\right)\)
b, \(y=\frac{1}{1-cosx}+\frac{1}{1+cosx},x\in\left(0,\frac{\pi}{2}\right)\)
c, \(y=2+tan^2x+cot^2x+\frac{1}{sin^4x+cos^4x},x\in\left(0,\frac{\pi}{2}\right)\)
a/
\(y=\frac{1}{sinx}+\frac{1}{cosx}\ge\frac{4}{sinx+cosx}=\frac{4}{\sqrt{2}sin\left(x+\frac{\pi}{4}\right)}\ge\frac{4}{\sqrt{2}}=2\sqrt{2}\)
\(y_{min}=2\sqrt{2}\) khi \(\left\{{}\begin{matrix}sinx=cosx\\sin\left(x+\frac{\pi}{4}\right)=1\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{4}\)
\(y_{max}\) không tồn tại (y dần tới dương vô cùng khi x gần tới 0 hoặc \(\frac{\pi}{2}\))
b/
\(y=\frac{1}{1-cosx}+\frac{1}{1+cosx}=\frac{1+cosx+1-cosx}{1-cos^2x}=\frac{2}{sin^2x}\)
Hàm số ko tồn tại cả min lẫn max ( \(0< y< \infty\))
c/
Do \(tan^2x\) ko tồn tại max (tiến tới vô cực) trên khoảng đã cho nên hàm ko tồn tại max
\(y=2+\frac{sin^4x+cos^4x}{\left(sinx.cosx\right)^2}+\frac{1}{sin^4x+cos^4x}\ge2+2\sqrt{\frac{sin^4x+cos^4x}{\frac{1}{4}sin^22x.\left(sin^4x+cos^4x\right)}}\)
\(y\ge2+\frac{4}{sin2x}\ge2+\frac{4}{1}=6\)
\(y_{min}=6\) khi \(\left\{{}\begin{matrix}sin2x=1\\sin^4x+cos^4x=sinx.cosx\end{matrix}\right.\) \(\Rightarrow x=\frac{\pi}{4}\)
giải các phương trình sau:
a, \(\sqrt{3}sinx+cosx=\frac{1}{cosx}\)
b,\(3tan^2x\left(x-\frac{\pi}{2}\right)=2\left(\frac{1-sinx}{sinx}\right)\)
c,\(1+sinx+cosx+tanx=0\)
d,\(\frac{1}{cosx}+\frac{1}{sinx}=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
\(cos^2x+\frac{1}{cos^2x}+cosx-\frac{1}{cosx}-\frac{7}{4}=0\)
mọi người ơi giúp em với :((em cảm ơn mọi người nhìu
ĐKXĐ: ...
Đặt \(cosx-\frac{1}{cosx}=a\Rightarrow cos^2x+\frac{1}{cos^2x}=a^2+2\)
Pt trở thành:
\(a^2+2+a-\frac{7}{4}=0\)
\(\Leftrightarrow4a^2+4a+1=0\Leftrightarrow\left(2a+1\right)^2=0\)
\(\Rightarrow a=-\frac{1}{2}\Rightarrow cosx-\frac{1}{cosx}=-\frac{1}{2}\)
\(\Leftrightarrow2cos^2x+cosx-2=0\)
\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{17}-1}{4}\\cosx=\frac{-\sqrt{17}-1}{4}< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=\pm arccos\left(\frac{\sqrt{17}-1}{4}\right)+k2\pi\)
giải các pt
a) \(sinx+cosx-2sin2x-1=0\)
b) \(sinx+cosx+3sinx.cosx-1=0\)
c) \(sinx-2sin2x=\frac{1}{2}-cosx\)
d) \(6\left(sinx-cosx\right)-1=sinx.cosx\)
a/
\(\Leftrightarrow sinx+cosx-4sinx.cosx-1=0\)
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\Rightarrow\left|t\right|\le\sqrt{2}\)
\(\Rightarrow t^2=1+2sinx.cosx\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)
Pt trở thành:
\(t-2\left(t^2-1\right)-1=0\)
\(\Leftrightarrow-2t^2+t+1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1\\\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\frac{1}{2\sqrt{2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\\x+\frac{\pi}{4}=arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(-\frac{1}{2\sqrt{2}}\right)+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\)
b/
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)
Pt trở thành:
\(t+\frac{3}{2}\left(t^2-1\right)-1=0\)
\(\Leftrightarrow3t^2+2t-5=0\)
\(\Rightarrow\left[{}\begin{matrix}t=-1\\t=\frac{5}{3}>\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow sinx+cosx-4sinx.cosx=\frac{1}{2}\)
Đặt \(sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=t\) với \(\left|t\right|\le\sqrt{2}\)
\(sinx.cosx=\frac{t^2-1}{2}\)
Pt trở thành:
\(t-2\left(t^2-1\right)=\frac{1}{2}\)
\(\Leftrightarrow-4t^2+2t+3=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{1+\sqrt{13}}{4}\\t=\frac{1-\sqrt{13}}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{1+\sqrt{13}}{4\sqrt{2}}\\sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{13}}{4\sqrt{2}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=arcsin\left(\frac{1+\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(\frac{1+\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=arcsin\left(\frac{1-\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\\x+\frac{\pi}{4}=\pi-arcsin\left(\frac{1-\sqrt{13}}{4\sqrt{2}}\right)+k2\pi\end{matrix}\right.\)
\(\Rightarrow x=...\)
giải phương trình :
\(\frac{sin^2x-2sinx-cosx+1}{cosx}=0\)
ĐKXĐ: \(cosx\ne0\)
Đặt \(\left\{{}\begin{matrix}sinx=a\\cosx=b\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=1\\a^2-2a-b+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=1\\\left(a-1\right)^2=b\end{matrix}\right.\) \(\Rightarrow a^2+\left(a-1\right)^4=1\)
\(\Leftrightarrow a^4-4a^3+7a^2-4a=0\)
\(\Leftrightarrow a\left(a-1\right)\left(a^2-3a+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\Rightarrow b=1\\a=1\Rightarrow b=0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=k2\pi\)
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.
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)