Giải pt: 3tan2x+\(\frac{3\left(tanx+1\right)}{Cosx}\)=1+\(4\sqrt{2}sin\left(x-\frac{15\pi}{4}\right)\)
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) \(\left(1+tanx\right)sin^2x=3sinx\left(cosx-sinx\right)+3\)
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,\(\frac{1}{sinx}+\frac{1}{sin\left(x-\frac{3\pi}{2}\right)}=4sin\left(\frac{7\pi}{4}-x\right)\)
Giair các pt lượng giác sau:
1) \(sin\left(x-\frac{\pi}{4}\right)\left(2cos+\sqrt{2}\right)tan2x=0\)
2) \(tan2x.sinx+3\left(sin-\sqrt{3}tan2x\right)-3\sqrt{3}=0\)
3) \(\frac{cos2x}{sin\left(x+\frac{3\pi}{4}\right)}=\frac{sin\left(x+\frac{3\pi}{4}\right)}{cos2x}\)
4) \(\left(\frac{tanx-1}{tanx+1}+cot2x\right)\left(3tan-\sqrt{3}\right)=0;0< x< \pi\)
a/ ĐKXĐ: \(cos2x\ne0\)
\(\Leftrightarrow2x\ne\frac{\pi}{2}+k\pi\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
Pt tương đương:
\(\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\2cosx+\sqrt{2}=0\\sin2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\cosx=cos\left(\frac{3\pi}{4}\right)\\2x=k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=\frac{3\pi}{4}+k2\pi\left(l\right)\\x=-\frac{3\pi}{4}+k2\pi\left(l\right)\\x=\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x=\frac{k\pi}{2}\)
b/
ĐKXĐ: \(x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
\(\Leftrightarrow tan2x.sinx+3sinx-\sqrt{3}tan2x-3\sqrt{3}=0\)
\(\Leftrightarrow sinx\left(tan2x+3\right)-\sqrt{3}\left(tan2x+3\right)=0\)
\(\Leftrightarrow\left(sinx-\sqrt{3}\right)\left(tan2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=\sqrt{3}>1\left(vn\right)\\tan2x=-3\end{matrix}\right.\)
\(\Rightarrow2x=arctan\left(-3\right)+k\pi\)
\(\Rightarrow x=\frac{arctan\left(-2\right)}{2}+\frac{k\pi}{2}\)
c/
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(x+\frac{3\pi}{4}\right)\ne0\\cos2x\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+\frac{3\pi}{4}\ne k\pi\\2x\ne\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\ne-\frac{3\pi}{4}+k\pi\\x\ne\frac{\pi}{4}+\frac{k\pi}{2}\end{matrix}\right.\) \(\Rightarrow x\ne\frac{\pi}{4}+\frac{k\pi}{2}\)
Pt tương đương:
\(cos^22x=sin^2\left(x+\frac{3\pi}{4}\right)\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos4x=\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{3\pi}{2}\right)\)
\(\Leftrightarrow cos4x=-cos\left(2x+\frac{3\pi}{2}\right)=cos\left(2x+\frac{\pi}{2}\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=2x+\frac{\pi}{2}+k2\pi\\4x=-2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\left(l\right)\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)
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:
\(\left(1+tanx\right)tan^2x=3tanx\left(1-tanx\right)+3\left(1+tan^2x\right)\)
\(\Leftrightarrow tan^3x+tan^2x=3tanx-3tan^2x+3+3tan^2x\)
\(\Leftrightarrow tan^3x+tan^2x-3tanx-3=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(tan^2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\sqrt{3}\\tanx=-\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\frac{\pi}{3}+k\pi\\x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
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,\(\frac{1}{cosx}+\frac{1}{sinx}=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
giải pt
a) \(\sqrt{3}sinx+cosx=2\)
b) \(sin\left(\frac{\pi}{4}-2x\right)+sin\left(\frac{\pi}{4}+x\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=1\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=1\)
\(\Leftrightarrow x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{3}+k2\pi\)
b.
\(\sqrt{2}sin\left(\frac{\pi}{4}-2x\right)+\sqrt{2}sin\left(\frac{\pi}{4}+x\right)=1\)
\(\Leftrightarrow cos2x-sin2x+sinx+cosx=1\)
\(\Leftrightarrow1-2sin^2x-2sinx.cosx+sinx+cosx=1\)
\(\Leftrightarrow-2sinx\left(sinx+cosx\right)+sinx+cosx=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1-2sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\sinx=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow x=...\)
Giải phương trình
1.\(\frac{\left(2sinx+1\right).\left(3cos4x+2sinx\right)+4cos^2x+1}{1+sinx}=8\)
2.\(\frac{\left(1+sinx+cos2x\right).sin\left(x+\frac{\pi}{4}\right)}{1+tanx}=\frac{1}{\sqrt{2}}cosx\)
a/ ĐKXĐ: \(sinx\ne-1\)
\(\Leftrightarrow\left(2sinx+1\right)\left(3cos4x+2sinx\right)+4cos^2x+1=8+8sinx\)
\(\Leftrightarrow6sinx.cos4x+4sin^2x+3cos4x+2sinx+4cos^2x+1=8+8sinx\)
\(\Leftrightarrow6sinx.cos4x+3cos4x-6sinx-3=0\)
\(\Leftrightarrow6sinx\left(cos4x-1\right)+3\left(cos4x-1\right)=0\)
\(\Leftrightarrow\left(6sinx+3\right)\left(cos4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\cos4x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\1-2sin^22x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1-sin^2x\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sin^2x\left(1+sinx\right)\left(1-sinx\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=0\\sinx=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\\x=k\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}tanx\ne-1\\cosx\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left(1+sinx+cos2x\right).\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=cosx\left(1+\frac{sinx}{cosx}\right)\)
\(\Leftrightarrow\left(1+sinx+cos2x\right)\left(sinx+cosx\right)=cosx+sinx\)
\(\Leftrightarrow\left(cosx+sinx\right)\left(sinx+cos2x\right)=0\)
\(\Leftrightarrow sinx+cos2x=0\)
\(\Leftrightarrow-2sin^2x+sinx+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\left(l\right)\\sinx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
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:
\(2\left(a^2+4\right)+9a-1=0\)
\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
giải các pt
a) \(tanx-\frac{\sqrt{2}}{cosx}=1\)
b) \(\frac{2sinx-1}{cos4x}+\frac{2sinx-1}{sin4x-1}=0\)
c) \(sin\left(x+\frac{\pi}{4}\right)-cos\left(x-\frac{\pi}{4}\right)=1\)
d) \(\frac{sin2x-2cos2x-5}{2sin2x-cos2x-6}=0\)
a/ ĐKXĐ:...
\(\Leftrightarrow\frac{sinx}{cosx}-\frac{\sqrt{2}}{cosx}=1\)
\(\Leftrightarrow sinx-\sqrt{2}=cosx\)
\(\Leftrightarrow sinx-cosx=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow x-\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{3\pi}{4}+k2\pi\)
b/
ĐKXĐ: ...
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x-1\right)+cos4x\left(2sinx-1\right)=0\)
\(\Leftrightarrow2sinx.sin4x-2sinx-sin4x+1+2sinx.cos4x-cos4x=0\)
\(\Leftrightarrow2sinx\left(sin4x+cos4x\right)-\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x\right)-\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sin4x+cos4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin4x+cos4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(4x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\4x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\4x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\\x=\frac{k\pi}{2}\\x=\frac{\pi}{8}+\frac{k\pi}{2}\left(l\right)\end{matrix}\right.\)
c/
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sinx=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k2\pi\\x=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
d/
\(\Leftrightarrow sin2x-2cos2x-5=2sin2x-cos2x-6\)
\(\Leftrightarrow sin2x+cos2x=1\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
c/
Hình như câu này đề sai
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)-\sqrt{2}cos\left(x-\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sinx+cosx-\left(sinx+cosx\right)=\sqrt{2}\)
\(\Leftrightarrow0=\sqrt{2}\)
Pt vô nghiệm
d/ Hình như câu này đề cũng sai
\(\Leftrightarrow sin2x-2cos2x-5=0\)
\(\Leftrightarrow\frac{1}{\sqrt{5}}sin2x-\frac{2}{\sqrt{5}}cos2x=\sqrt{5}\)
\(\Leftrightarrow sin\left(2x-a\right)=\sqrt{5}\) (với \(sina=\frac{2}{\sqrt{5}};cosa=\frac{1}{\sqrt{5}}\))
Pt vô nghiệm do \(\sqrt{5}>1\)