Giải pt:
a) \(sinx+\frac{1}{sin2x}=cosx+\frac{1}{cosx}\)
b) \(sin^2x+\frac{sin2x}{2}=1\)
c) \(\frac{1}{1+cos^2x}+\frac{1}{1+sin^2x}=\frac{16}{11}\)
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: \(x\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)
\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)
\(\Leftrightarrow sin^3x-sin^2x=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=1\left(l\right)\end{matrix}\right.\)
\(\frac{2sin5x}{sinx+cosx}+\frac{1}{2}sin2x=sin^4x+cos^4x+sin^2x.cos^2x\)
giải các pt
a) \(sin\left(\frac{3\pi}{10}-\frac{x}{2}\right)=\frac{1}{2}sin\left(\frac{\pi}{10}+\frac{3x}{2}\right)\)
b) \(4\left(sin^2x+\frac{1}{sin^2x}\right)+4\left(sinx+\frac{1}{sinx}\right)=7\)
c) \(9\left(\frac{2}{cosx}+cosx\right)+2\left(cos^2x+\frac{4}{cos^2x}\right)=1\)
d) \(2\left(cos^2x+\frac{4}{cos^2x}\right)+9\left(\frac{2}{cosx}-cosx\right)=1\)
a/
\(\Leftrightarrow cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}sin\left(\frac{3x}{2}+\frac{\pi}{10}\right)\)
Đặt \(\frac{x}{2}+\frac{\pi}{5}=a\Rightarrow\frac{x}{2}=a-\frac{\pi}{5}\Rightarrow\frac{3x}{2}=3a-\frac{3\pi}{5}\)
Pt trở thành:
\(cosa=\frac{1}{2}sin\left(3a-\frac{3\pi}{5}+\frac{\pi}{10}\right)\)
\(\Leftrightarrow cosa=\frac{1}{2}sin\left(3a-\frac{\pi}{2}\right)\)
\(\Leftrightarrow cosa=-\frac{1}{2}sin\left(\frac{\pi}{2}-3a\right)=-\frac{1}{2}cos3a\)
\(\Leftrightarrow cos3a+2cosa=0\)
\(\Leftrightarrow4cos^3a-3cosa+2cosa=0\)
\(\Leftrightarrow4cos^3a-cosa=0\)
\(\Leftrightarrow cosa\left(4cos^2a-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}cosa=0\\cosa=\frac{1}{2}\\cosa=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=0\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=\frac{1}{2}\\cos\left(\frac{x}{2}+\frac{\pi}{5}\right)=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}+\frac{\pi}{5}=\frac{\pi}{2}+k\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{\pi}{3}+k2\pi\\\frac{x}{2}+\frac{\pi}{5}=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=...\) (5 nghiệm bạn tự biến đổi)
b/
ĐKXĐ: ...
Đặt \(sinx+\frac{1}{sinx}=a\Rightarrow sin^2x+\frac{1}{sin^2x}=a^2-2\)
Pt trở thành:
\(4\left(a^2-2\right)+4a=7\)
\(\Leftrightarrow4a^2+4a-15=0\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx+\frac{1}{sinx}=\frac{3}{2}\\sinx+\frac{1}{sinx}=-\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-\frac{3}{2}sinx+1=0\left(vn\right)\\sin^2x+\frac{5}{2}sinx+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
c/
ĐKXĐ: ...
Đặt \(cosx+\frac{2}{cosx}=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2-4\)
Pt trở thành:
\(9a+2\left(a^2-4\right)=1\)
\(\Leftrightarrow2a^2+9a-9=0\)
Pt này nghiệm xấu quá bạn :(
d/ĐKXĐ: ...
Đặt \(\frac{2}{cosx}-cosx=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2+4\)
Pt trở thành:
\(2\left(a^2+4\right)+9a-1=0\)
\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
giai pt
a) \(sin^6x+cos^6x+sin2x=1\)
b) \(1+sinx+cosx+sin2x+cos2x=0\)
c) \(sin^2x-3cos^2x=2\left(sinx+\sqrt{3}cosx\right)\)
d) \(sin^3x.cosx-sinx.cos^3x=\frac{\sqrt{2}}{8}\)
a/
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)+sin2x-1=0\)
\(\Leftrightarrow1-3sin^2x.cos^2x+sin2x-1=0\)
\(\Leftrightarrow-\frac{3}{4}sin^22x+sin2x=0\)
\(\Leftrightarrow sin2x\left(1-\frac{3}{4}sin2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin2x=\frac{4}{3}>1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow2x=k\pi\)
\(\Leftrightarrow x=\frac{k\pi}{2}\)
b/
\(\Leftrightarrow\left(1+sin2x\right)+sinx+cosx+cos^2x-sin^2x=0\)
\(\Leftrightarrow\left(sinx+cosx\right)^2+sinx+cosx+\left(sinx+cosx\right)\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+cosx+1+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\2cosx+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c/
\(\Leftrightarrow\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}\right)cosx=2\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}cosx=0\\sinx-\sqrt{3}cosx=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=0\\\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=0\\sin\left(x-\frac{\pi}{3}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=k\pi\\x-\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
1/Chứng minh rằng :
a/ cot\(^2\)x \(-cos^2x=cot^2x.cos^2x\)
b/ \(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=2tan2x\)
c/ \(\frac{sin4x+cos2x}{1+sin2x-cos4x}=cot2x\)
2/ Rút gọn biểu thức
A=\(sin^3+sin^2xcosx+sinxcos^2x+cos^3x\)
B=\(tanx\left(\frac{1+cos^2x}{sinx}-sinx\right)\)
\(cot^2x-cos^2x=\frac{cos^2x}{sin^2x}-cos^2x=cos^2x\left(\frac{1}{sin^2x}-1\right)=\frac{cos^2x\left(1-sin^2x\right)}{sin^2x}\)
\(=cos^2x.\left(\frac{cos^2x}{sin^2x}\right)=cot^2x.cos^2x\)
\(\frac{cosx+sinx}{cosx-sinx}-\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx+sinx\right)^2-\left(cosx-sinx\right)^2}{\left(cosx-sinx\right)\left(cosx+sinx\right)}\)
\(=\frac{cos^2x+sin^2x+2sinx.cosx-\left(cos^2x+sin^2x-2sinx.cosx\right)}{cos^2x-sin^2x}=\frac{4sinx.cosx}{cos2x}=\frac{2sin2x}{cos2x}=2tan2x\)
\(\frac{sin4x+cos2x}{1-cos4x+sin2x}=\frac{2sin2x.cos2x+cos2x}{1-\left(1-2sin^22x\right)+sin2x}=\frac{cos2x\left(2sin2x+1\right)}{sin2x\left(2sin2x+1\right)}=\frac{cos2x}{sin2x}=cot2x\)
\(A=sin^2x\left(sinx+cosx\right)+cos^2x\left(sinx+cosx\right)\)
\(=\left(sin^2x+cos^2x\right)\left(sinx+cosx\right)=sinx+cosx\)
\(B=\frac{sinx}{cosx}\left(\frac{1+cos^2x-sin^2x}{sinx}\right)=\frac{sinx}{cosx}\left(\frac{2cos^2x}{sinx}\right)=2cosx\)
giải phương trình
1.\(sin^3x+2cosx-2+sin^2x=0\)
\(2.\frac{\sqrt{3}}{2}sin2x+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
3.\(2sin2x-cos2x=7sinx+2cosx-4\)
4.\(2cos2x-8cosx+7=\frac{1}{cosx}\)
5.\(cos^8x+sin^8x=2\left(cos^{10}x+sin^{10}x\right)+\frac{5}{4}cos2x\)
6.\(1+sinx+cos3x=cosx+sin2x+cos2x\)
7.\(1+sinx+cosx+sin2x+cos2x=0\)
1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm
3.
\(\Leftrightarrow4sinx.cosx-\left(1-2sin^2x\right)=7sinx+2cosx-4\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+2sin^2x-7sinx+3=0\)
\(\Leftrightarrow2cosx\left(2sinx-1\right)+\left(sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2cosx+sinx-3\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\Leftrightarrow...\\2cosx+sinx=3\left(1\right)\end{matrix}\right.\)
Xét (1), do \(2^2+1^2< 3^2\) nên (1) vô nghiệm
a) \(1-cot^4x=\frac{2}{sin^2x}-\frac{1}{sin^4x}\)
b)\(\frac{1-2sinx.cosx}{cos^2-sin^2}\)\(=\frac{1-tanx}{1+tanx}\)\(\)
c)\(\frac{sin^2x}{sinx-cosx}+\frac{sinx+cosx}{1-tanx}=sinx+cosx\)
d)\(\sqrt{\frac{1+cosx}{1-cosx}}-\sqrt{\frac{1-cosx}{1+cosx}}=\frac{2.cosx}{|sin|}\)
e)\(tan^3x+tan^2x+tanx+1=\frac{sinx+cosx}{cos^3x}\)
Chứng minh:
a) \(tan(\frac\pi4+\frac{x}2).\frac{1+cos(\frac\pi2+x)}{sin(\frac\pi2+x)}=1\)
b) \(tan(\frac\pi4+x)=\frac{1+sin2x}{cos2x}\)
c) \(\frac{cosx}{1-sinx}=cot(\frac\pi4-\frac{x}{2})\)
d) \(tanx.tan3x=\frac{tan^22x-tan^2x}{1-tan^2x.tan^22x}\)
Chứng minh|
a) \(\frac{1+sin2x}{sinx+cosx}-\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}=sinx\)
b) \(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)
\(\frac{sin^2x+cos^2x+2sinx.cosx}{sinx+cosx}-\left(1-tan^2\frac{x}{2}\right).cos^2\frac{x}{2}\)
\(=\frac{\left(sinx+cosx\right)^2}{sinx+cosx}-\left(cos^2\frac{x}{2}-sin^2\frac{x}{2}\right)\)
\(=sinx+cosx-cosx=sinx\)
\(sin^4x+cos^4\left(x+\frac{\pi}{4}\right)=\left(\frac{1}{2}-\frac{1}{2}cos2x\right)^2+\left(\frac{1}{2}+\frac{1}{2}cos\left(2x+\frac{\pi}{2}\right)\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\left(\frac{1}{2}-\frac{1}{2}sin2x\right)^2\)
\(=\frac{1}{4}-\frac{1}{2}cos2x+\frac{1}{4}cos^22x+\frac{1}{4}-\frac{1}{2}sin2x+\frac{1}{4}sin^22x\)
\(=\frac{1}{4}-\frac{1}{2}\left(cos2x+sin2x\right)+\frac{1}{4}\left(cos^22x+sin^22x\right)\)
\(=\frac{3}{4}-\frac{\sqrt{2}}{2}sin\left(2x+\frac{\pi}{4}\right)\)