giải phương trình:
a, \(tanx.sin^2x-2sin^2x=3\left(cos2x+sinxcosx\right)\)
b, \(5sinx-2=3\left(1-sinx\right)tan^2x\)
c,\(\frac{cos2x+3cot2x+4sinx}{cot2x-cos2x}=2\)
d, \(\frac{4sin^2x+6sin^2x-3cos2x-9}{cosx}=0\)
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10 tháng 10 2021 lúc 16:04
Giải các pt sau
a, \(\dfrac{1}{sinx}+\dfrac{1}{cosx}=4sin\left(x+\dfrac{\pi}{4}\right)\)
b, \(2sin\left(2x-\dfrac{\pi}{6}\right)+4sinx+1=0\)
c, \(cos2x+\sqrt{3}sinx+\sqrt{3}sin2x-cosx=2\)
d, \(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+cos^2\left(x-\dfrac{3\pi}{4}\right)\)
giải phương trình:
1) \(2\sqrt{2}cos^3x\left(x-\frac{\pi}{4}\right)-3cosx-sinx=0\)
2) \(tanx.sin^2x-2sin^2x=3\left(cos2x+sinxcosx\right)\)
3) \(2sin^3x=cosx\)
4) \(6sinx-2cos^3x=\frac{5sin4xcosx}{2cos2x}\)
Giai pt
\(5sinx-2=3\left(1-sinx\right)tan^2x\)
\(2.cos2x.cosx=1+cos2x+cos3x\)
\(cos2x+cosx=4sin^2\left(\dfrac{x}{2}\right)-1\)
Bạn tham khảo pt 1 hộ mình nha. Chúc bạn học tốt~
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Giai pt
\(5sinx-2=3\left(1-sinx\right)tan^2x\)
\(2.cos2x.cosx=1+cos2x+cos3x\)
\(cos2x+cosx+4sin^2\left(\dfrac{x}{2}\right)-1\)
Pt 1.
Bạn tham khảo phương trình 1 hộ mình nha. Chúc bạn học tốt
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25 tháng 7 2020 lúc 15:32
giải các pt
a) \(\left(2sin^2x-1\right)tan^22x+3\left(2cos^2x-1\right)=0\)
b) \(tanx+tan2x=\frac{2sin3x}{sin2x}\)
c) \(1+sinx.cos2x=sinx+cos2x\)
d) \(tanx=1-cos2x\)
a/
DKXD: ...
\(\Leftrightarrow-cos2x.tan^22x+3.cos2x=0\)
\(\Leftrightarrow cos2x\left(3-tan^22x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\tan^22x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}+k\pi\\tan2x=\sqrt{3}\\tan2x=-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\2x=\frac{\pi}{3}+k\pi\\2x=-\frac{\pi}{3}+k\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=\pm\frac{\pi}{6}+\frac{k\pi}{2}\end{matrix}\right.\)
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b/
DKXD: ...
\(\Leftrightarrow\frac{sinx}{cosx}+\frac{sin2x}{cos2x}-\frac{2sin3x}{sin2x}=0\)
\(\Leftrightarrow\frac{sinx.cos2x+sin2x.cosx}{cosx.cos2x}-\frac{2sin3x}{sin2x}=0\)
\(\Leftrightarrow\frac{sin\left(2x+x\right)}{cosx.cos2x}-\frac{2sin3x}{sin2x}=0\)
\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}-\frac{2sin3x}{sin2x}=0\)
\(\Leftrightarrow sin3x\left(\frac{1}{cosx.cos2x}-\frac{2}{sin2x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\left(1\right)\\2cosx.cos2x=sin2x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3sinx-4sin^3x=0\) (tìm nghiệm thẳng bằng \(3x=k\pi\) rồi dựa vào đường tròn lượng giác loại nghiệm cũng được)
\(\Leftrightarrow sinx\left(3-4sin^2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(l\right)\\sinx=\pm\frac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=\frac{2\pi}{3}+k2\pi\\x=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow2cosx.cos2x=2sinx.cosx\)
\(\Leftrightarrow2cosx\left(cos2x-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\left(l\right)\\cos2x=sinx=cos\left(\frac{\pi}{2}-x\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{2}-x+k2\pi\\2x=x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+\frac{k2\pi}{3}\\x=-\frac{\pi}{2}+k2\pi\left(l\right)\end{matrix}\right.\)
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c/
\(\Leftrightarrow sinx.cos2x-sinx+1-cos2x=0\)
\(\Leftrightarrow sinx\left(cos2x-1\right)-\left(cos2x-1\right)=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(cos2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\cos2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\2x=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=k\pi\end{matrix}\right.\)
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6 tháng 12 2016 lúc 22:44
giải các phương trình sau :a) sinleft(x-frac{2pi}{3}right)cos2x ; b) tanleft(2x+45^oright)tanleft(180^o-frac{x}{2}right)1 ; c) cos2x-sin^2x0 ; d) 5tan x-2cot x3 ; e) sin2x+sin^2xfrac{1}{2} ; f) sin^2frac{x}{2}+sin x-2cos^2frac{x}{2}frac{1}{2} ; g) frac{1+cos2x}{cos x}frac{sin2x}{1-cos2x}
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giải các phương trình sau :
a) \(\sin\left(x-\frac{2\pi}{3}\right)=\cos2x\) ; b) \(\tan\left(2x+45^o\right)\tan\left(180^o-\frac{x}{2}\right)=1\) ; c) \(\cos2x-\sin^2x=0\) ; d) \(5\tan x-2\cot x=3\) ; e)
\(\sin2x+\sin^2x=\frac{1}{2}\) ; f) \(\sin^2\frac{x}{2}+\sin x-2\cos^2\frac{x}{2}=\frac{1}{2}\) ; g) \(\frac{1+\cos2x}{\cos x}=\frac{\sin2x}{1-\cos2x}\)
mai đăng lại bài này nhé t làm cho h đi ngủ
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Giải các phương trình lượng giác sau:1) a/ cosleft(10x+12right)+4sqrt{2}sinleft(5x+6right)-40 b/ cosleft(4x+2right)+3sinleft(2x+1right)22) a/ cos2x+sin^2x+2cosx+10 b/ 4sin^22x-8cos^2x+ 30 c/ 4cos2x+4sin^2x+4sinx13) a/ tanx+cotx2 b/ 2tanx-2cotx34) a/ 2sin2x+8tanx9sqrt{3} b/ 2cos2x+tan^2x55) a/ left(3+cotxright)^25left(3+cotxright) b/ 4left(sin^2x+dfrac{1}{sin^2x}right)-4left(sinx+dfrac{1}{sinx}right)7
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Giải các phương trình lượng giác sau:
1) a/ \(cos\left(10x+12\right)+4\sqrt{2}sin\left(5x+6\right)-4=0\)
b/ \(cos\left(4x+2\right)+3sin\left(2x+1\right)=2\)
2) a/ \(cos2x+sin^2x+2cosx+1=0\)
b/ \(4sin^22x-8cos^2x+ 3=0\)
c/ \(4cos2x+4sin^2x+4sinx=1\)
3) a/ \(tanx+cotx=2\)
b/ \(2tanx-2cotx=3\)
4) a/ \(2sin2x+8tanx=9\sqrt{3}\)
b/ \(2cos2x+tan^2x=5\)
5) a/ \(\left(3+cotx\right)^2=5\left(3+cotx\right)\)
b/ \(4\left(sin^2x+\dfrac{1}{sin^2x}\right)-4\left(sinx+\dfrac{1}{sinx}\right)=7\)
1a.
Đặt \(5x+6=u\)
\(cos2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow1-2sin^2u+4\sqrt{2}sinu-4=0\)
\(\Leftrightarrow2sin^2u-4\sqrt{2}sinu+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=\dfrac{3\sqrt{2}}{2}>1\left(loại\right)\\sinu=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow sin\left(5x+6\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+6=\dfrac{\pi}{4}+k2\pi\\5x+6=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}+\dfrac{\pi}{20}+\dfrac{k2\pi}{5}\\x=-\dfrac{6}{5}+\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
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1b.
Đặt \(2x+1=u\)
\(cos2u+3sinu=2\)
\(\Leftrightarrow1-2sin^2u+3sinu=2\)
\(\Leftrightarrow2sin^2u-3sinu+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinu=1\\sinu=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x+1\right)=1\\sin\left(2x+1\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{\pi}{2}+k2\pi\\2x+1=\dfrac{\pi}{6}+k2\pi\\2x+1=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}+\dfrac{\pi}{4}+k\pi\\x=-\dfrac{1}{2}+\dfrac{\pi}{12}+k\pi\\x=-\dfrac{1}{2}+\dfrac{5\pi}{12}+k\pi\end{matrix}\right.\)
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2a.
\(cos^2x-sin^2x+sin^2x+2cosx+1=0\)
\(\Leftrightarrow cos^2x+2cosx+1=0\)
\(\Leftrightarrow\left(cosx+1\right)^2=0\)
\(\Leftrightarrow cosx=-1\)
\(\Leftrightarrow x=\pi+k2\pi\)
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27 tháng 8 2020 lúc 16:50
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/
\(\Leftrightarrow4sinx.cosx\left(sin^4x-cos^4x\right)=sin^24x\)
\(\Leftrightarrow2sin2x\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^24x\)
\(\Leftrightarrow-2sin2x.cos2x=sin^24x\)
\(\Leftrightarrow-sin4x=sin^24x\)
\(\Leftrightarrow\left[{}\begin{matrix}sin4x=0\\sin4x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=k\pi\\4x=-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{k\pi}{4}\\x=-\frac{\pi}{8}+\frac{k\pi}{2}\end{matrix}\right.\)
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b/
\(\Leftrightarrow2\left(1-cosx\right)-\sqrt{3}cos2x=1+1+cos\left(2x-\frac{3\pi}{2}\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=sin\left(2\pi-2x\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)
\(\Leftrightarrow sin2x-\sqrt{3}cos2x=2cosx\)
\(\Leftrightarrow\frac{1}{2}sin2x-\sqrt{3}cos2x=cosx\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=cosx=sin\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\frac{\pi}{3}=\frac{\pi}{2}-x+k2\pi\\2x-\frac{\pi}{3}=\frac{\pi}{2}+x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{18}+\frac{k2\pi}{3}\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
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c/
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)-\frac{5}{4}=0\)
\(\Leftrightarrow sin^2\left(x+\frac{\pi}{3}\right)+2sin\left(x+\frac{\pi}{3}\right)-\frac{5}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\\sin\left(x+\frac{\pi}{3}\right)=-\frac{5}{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x+\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
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1 tháng 10 2021 lúc 9:12
a)4sin^3xcos3x+4cos^3xsin3x+3sqrt{3}cos4x3b)2sin^2xleft(4sin^4x-1right)cos2xleft(7cos^22x+3cos2x-4right)2 câu này giải như nào ạ
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a)\(4sin^3xcos3x+4cos^3xsin3x+3\sqrt{3}cos4x=3\)
b)\(2sin^2x\left(4sin^4x-1\right)=cos2x\left(7cos^22x+3cos2x-4\right)\)
2 câu này giải như nào ạ
a
\(\Leftrightarrow\left(3sinx-sin3x\right)cos3x+\left(3cosx+cos3x\right)sin3x+3\sqrt{3}cos4x=3\)
\(\Leftrightarrow\left(sinx.cos3x+sin3x.cosx\right)+\sqrt{3}cos4x=1\)
\(\Leftrightarrow sin4x+\sqrt{3}cos4x=1\)
Tới đây thôi, mình lười ghi rồi =))
b
\(\Leftrightarrow\left(1-cos2x\right)\left(2sin^2x-1\right)\left(2sin^2+1\right)=cos2x\left(7cos^22x+3cos2x-4\right)\)
\(\Leftrightarrow\left(1-cos2x\right)\left(-cos2x\right)\left(2-cos2x\right)=cos2x\left(7cos^22x+3cos2x+4\right)\)
\(\Leftrightarrow-cos^22x+3cos2x-2=7cos^22x+3cos2x+4\)
\(\Leftrightarrow4cos^22x+3=0\)
=> pt vô nghiệm
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