Giải phương trình
1.\(\frac{\left(2sinx+1\right).\left(3cos4x+2sinx\right)+4cos^2x+1}{1+sinx}=8\)
2.\(\frac{\left(1+sinx+cos2x\right).sin\left(x+\frac{\pi}{4}\right)}{1+tanx}=\frac{1}{\sqrt{2}}cosx\)
giải phương trình sau:
a,\(\frac{sin2x+2cosx-sinx-1}{tanx+\sqrt{3}}=0\)
b,\(\frac{\left(1+sinx+cos2x\right)sinx\left(x+\frac{\pi}{4}\right)}{1+tanx}=\frac{1}{\sqrt{2}}cosx\)
c,\(\frac{\left(1-sin2x\right)cosx}{\left(1+sin2x\right)\left(1-sinx\right)}=\sqrt{3}\)
d,\(\frac{1}{sinx}+\frac{1}{sin\left(x-\frac{3\pi}{2}\right)}=4sin\left(\frac{7\pi}{4}-x\right)\)
giai cac pt
a) \(sin^3\left(x+\frac{\pi}{4}\right)=\sqrt{2}sinx\)
b) \(cos^3x-sin^3x=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)
c) \(\frac{1-tanx}{1+tanx}=1+2sinx\)
d) \(\left(1+tanx\right)sin^2x=3sinx\left(cosx-sinx\right)+3\)
giai pt:
a) \(4sin^5x.cosx-4cos^5x.sinx=sin^24x\)
b) \(4sin^2\frac{x}{2}-\sqrt{3}cos2x=1+2cos^2\left(x-\frac{3\pi}{4}\right)\)
c) \(sin^2\left(x+\frac{\pi}{3}\right)+sinx+\sqrt{3}cosx=\frac{5}{4}\)
d) \(2sinx\left(1+cos2x\right)+sin2x=1+2cosx\)
e) \(sin^2x+4sinx.cosx+3cos^2x-sinx-3ccosx=0\)
a/
\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)
\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)
\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)
\(\Leftrightarrow-sin4x=sin^24x\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=k\pi\\4x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
b/
\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)
\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)
\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{2}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
giai pt
a) \(cos^3x-sin^3x=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\)
b) \(\frac{1-tanx}{1+tanx}=1+2sinx\)
c) \(\left(1+tanx\right)sin^2x=3sinx\left(cosx-sinx\right)+3\)
a/
\(\Leftrightarrow cos^3x-sin^3x=cosx+sinx\)
- Với \(cosx=0\Rightarrow sinx=-1\Rightarrow x=-\frac{\pi}{2}+k2\pi\) là 1 nghiệm
- Với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow1-tan^3x=\frac{1}{cos^2x}+tanx.\frac{1}{cos^2x}\)
\(\Leftrightarrow1-tan^3x=1+tan^2x+tanx\left(1+tan^2x\right)\)
\(\Leftrightarrow2tan^3x+tan^2x+tanx=0\)
\(\Leftrightarrow tanx\left(2tan^2x+tanx+1\right)=0\)
\(\Leftrightarrow tanx=0\Rightarrow x=k\pi\)
b/
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\frac{1-\frac{sinx}{cosx}}{1+\frac{sinx}{cosx}}=1+2sinx\)
\(\Leftrightarrow\frac{cosx-sinx}{cosx+sinx}=1+2sinx\)
\(\Leftrightarrow cosx-sinx=\left(1+2sinx\right)\left(cosx+sinx\right)\)
\(\Leftrightarrow sinx+sinx.cosx+sin^2x=0\)
\(\Leftrightarrow sinx\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\sinx+cosx=-1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{4}=-\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\left(l\right)\\x=\pi+k2\pi\end{matrix}\right.\)
c/
ĐKXĐ: ...
Chia 2 vế cho \(cos^2x\) ta được:
\(\left(1+tanx\right)tan^2x=3tanx\left(1-tanx\right)+3\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^3x+tan^2x=3tanx-3tan^2x+3+3tan^2x\)
\(\Leftrightarrow tan^3x+tan^2x-3tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
\(cosx-2cos3x=1+\sqrt{3}sinx\)
\(sinx+sinx\left(x+\dfrac{\pi}{3}\right)+sin4x=sin\left(2x-\dfrac{\pi}{3}\right)\)
\(\left(1-\dfrac{1}{2sinx}\right)cos^22x=2sinx-3+\dfrac{1}{sinx}\)
( sinx -2cosx)cos2x + sinx = (cos4x - 1)cosx +\(\dfrac{cos2x}{2sinx}\)
\(\left(\dfrac{cos4x+sin2x}{cos3x+sin3x}\right)^2=2\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+3\)
giai pt:
a) \(\left(2cosx-1\right)\left(2sinx+cosx\right)=sin2x-sinx\)
b) \(\frac{sin2x}{cosx}+\frac{cos2x}{sinx}=tanx-cotx\)
c) \(\frac{1}{cos^2x}=\frac{2-sin^3x-cos^2x}{1-sin^3x}\)
a/
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sin2x.sinx+cos2x.cosx}{sinx.cosx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}\)
\(\Leftrightarrow\frac{cos\left(2x-x\right)}{sinx.cosx}=\frac{sin^2x-cos^2x}{sinx.cosx}\)
\(\Leftrightarrow cosx=sin^2x-cos^2x\)
\(\Leftrightarrow cosx=1-2cos^2x\)
\(\Leftrightarrow2cos^2x+cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(l\right)\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
c/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)
\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)
\(\Leftrightarrow sin^3x-sin^2x=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=1\left(l\right)\end{matrix}\right.\)
Giải pt sau :
1/ (2sinx-1)(2cos2x+2sinx+1)=3-4cos2 x
2/ \(\sqrt{3}cot\left(\frac{\pi}{4}-x\right)+1=0\)
3/ (cos\(\frac{x}{4}-3sinx\)) sinx + (\(\left(1+sin\frac{x}{4}-3cosx\right)cosx=0\)
4/ \(sin2x-cos2x+3sinx-cosx-1=0\)
1.
\(\left(2sinx-1\right)\left(2cos2x+2sinx+1\right)=3-4\left(1-sin^2x\right)\)
\(\Leftrightarrow\left(2sinx-1\right)\left(2cos2x+2sinx+1\right)=4sin^2x-1\)
\(\Leftrightarrow\left(2sinx-1\right)\left(2cos2x+2sinx+1\right)-\left(2sinx-1\right)\left(2sinx+1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(2cos2x+2sinx+1-2sinx-1\right)=0\)
\(\Leftrightarrow2cos2x\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
2.
ĐKXĐ: ...
\(\Leftrightarrow cot\left(\frac{\pi}{4}-x\right)=-\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow\frac{\pi}{4}-x=-\frac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\frac{7\pi}{12}+k\pi\)
3.
\(\Leftrightarrow cos\frac{x}{4}sinx+sin\frac{x}{4}.cosx-3\left(sin^2x+cos^2x\right)+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{x}{4}\right)=-cosx\)
\(\Leftrightarrow sin\frac{5x}{4}=sin\left(x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{5x}{4}=x-\frac{\pi}{2}+k2\pi\\\frac{5x}{4}=\frac{3\pi}{2}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
4.
\(\Leftrightarrow2sinx.cosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+2sin^2x+3sinx-2=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+\left(2sinx-1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\sinx+cosx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
Câu 1: Giải các phương trình sau:
a, \(\left(sin\frac{x}{2}+cos\frac{x}{2}\right)^2\)+\(\sqrt{3}cosx=2\)
b, \(\frac{\left(1-2sinx\right).cosx}{\left(1+2sinx\right)\left(1-sinx\right)}=\sqrt{3}\)
c, 5sinx-2=3(1-sinx).tan2x
d, \(\frac{2\left(sin^6x+cos^6\right)}{\sqrt{2}-2sinx}=0\)
e, cos23x.cos2x-cos2x=0
Câu 2: giải các phương trình sau:
a, sinx+cosx.sin2x+\(\sqrt{3}cos3x=2\left(cos4x+sin^3x\right)\)
b, \(\frac{\left(2-\sqrt{3}\right).cosx-2sin2\left(\frac{x}{2}-\frac{\pi}{4}\right)}{2cosx-1}\)
c, 8sin22x.cos2x=\(\sqrt{3}sin2x+cos2x\)
d, sin3x- \(\sqrt{3}cos^3x=sinxcos^2x-\sqrt{3}sin^2xcosx\)
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,\(\frac{1}{cosx}+\frac{1}{sinx}=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)