\(\frac{sin2x+2cosx-sinx-1}{tanx+\sqrt{3}}=0\)
Những câu hỏi liên quan
giải phương trình sau:
a,frac{sin2x+2cosx-sinx-1}{tanx+sqrt{3}}0
b,frac{left(1+sinx+cos2xright)sinxleft(x+frac{pi}{4}right)}{1+tanx}frac{1}{sqrt{2}}cosx
c,frac{left(1-sin2xright)cosx}{left(1+sin2xright)left(1-sinxright)}sqrt{3}
d,frac{1}{sinx}+frac{1}{sinleft(x-frac{3pi}{2}right)}4sinleft(frac{7pi}{4}-xright)
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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,\(\frac{1}{sinx}+\frac{1}{sin\left(x-\frac{3\pi}{2}\right)}=4sin\left(\frac{7\pi}{4}-x\right)\)
1> 1 + sinx + cosx + sin2x + cos2x = 0
2> cos2x + 3sin2x + 5 sinx - 3cosx = 3
3> \(\dfrac{\sqrt{2}*(cosx - sinx)}{cotx - 1}\) = \(\dfrac{1}{tanx + cot2x}\)
4> (2cosx - 1)*(2sinx + cosx) = sin2x - sinx
1)\(\dfrac{1+sin2x+cos2x}{1+cot^2x}\)=\(\sqrt{2}\)sinxsin2x
2)sin2xcosx+sinxcosx=cos2x+sinx+cosx
3)\(\dfrac{sin2x+2cosx-sinx-1}{tanx+\sqrt{3}}\)=0
Mấy bạn giải giúp mình nha,đang cần gấp
27 tháng 8 2020 lúc 17:07
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/
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)=2sinx.cosx-sinx\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx\right)-sinx\left(2cosx-1\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(2sinx+cosx-sinx\right)=0\)
\(\Leftrightarrow\left(2cosx-1\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2cosx-1=0\\sinx+cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{\pi}{3}+k2\pi\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
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b/ ĐKXĐ: \(x\ne\frac{k\pi}{2}\)
\(\Leftrightarrow\frac{sin2x.sinx+cos2x.cosx}{sinx.cosx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}\)
\(\Leftrightarrow\frac{cos\left(2x-x\right)}{sinx.cosx}=\frac{sin^2x-cos^2x}{sinx.cosx}\)
\(\Leftrightarrow cosx=sin^2x-cos^2x\)
\(\Leftrightarrow cosx=1-2cos^2x\)
\(\Leftrightarrow2cos^2x+cosx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(l\right)\\cosx=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=\pm\frac{\pi}{3}+k2\pi\)
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c/ ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{1-cos^2x+1-sin^3x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1}{cos^2x}=\frac{sin^2x}{1-sin^3x}+1\)
\(\Leftrightarrow\frac{1}{cos^2x}-1=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{1-cos^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\frac{sin^2x}{cos^2x}=\frac{sin^2x}{1-sin^3x}\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\Rightarrow x=k\pi\\cos^2x=1-sin^3x\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow1-sin^2x=1-sin^3x\)
\(\Leftrightarrow sin^3x-sin^2x=0\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=1\left(l\right)\end{matrix}\right.\)
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Giúp mình với mn...
1)cos2x+cos22x+cos23x+cos24x=2
2) (1-tanx) (1+sin2x)=1+tanx
3) tan2x=sin3x.cosx
4) tanx +cot2x=2cot4x
5) sinx+sin2x+sin3x=cosx+cos2x+cos3x
6)sinx=√2 sin5x-cosx
7) 1/sin2x + 1/cos2x =2/sin4x
8) sinx+cosx=cos2x/1-sin2x
9)1+cos2x/cosx= sin2x/1-cos2x
10)sin3x+cos3x/2cosx-sinx=cos2x
Giải phương trình :
Cotx=tanx +2cos4x\sin2x
Và
sinx +cos2x - 2cosx×cos3x=0
câu 1
⇒ \(\dfrac{cosx}{sinx}\) - \(\dfrac{sinx}{cosx}\) -\(\dfrac{2cos4x}{2sinxcosx}\) =0
⇔ \(\dfrac{cos^2x-sin^2x}{sinx.cosx}\) -\(\dfrac{cos4x}{sinx.cosx}\)= 0
⇔ \(\dfrac{cos2x-cos4x}{sinx.cosx}\) = 0
\(\left[{}\begin{matrix}cos2x=cos4x\\sin2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4x+k2\pi\\2x=-4x+k2\pi\\2x=k\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-k\pi\\x=\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\) (k∈ Z)
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câu 2 dùng công thức biến đổi tích thành tổng thành cos 4x + cos 2x sau đó phương trình trở thành sin x - cos 4x=0
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giải phương trình
1.sin^3x+2cosx-2+sin^2x0
2.frac{sqrt{3}}{2}sin2x+sqrt{2}cos^2x+sqrt{6}cosx0
3.2sin2x-cos2x7sinx+2cosx-4
4.2cos2x-8cosx+7frac{1}{cosx}
5.cos^8x+sin^8x2left(cos^{10}x+sin^{10}xright)+frac{5}{4}cos2x
6.1+sinx+cos3xcosx+sin2x+cos2x
7.1+sinx+cosx+sin2x+cos2x0
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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
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Xem thêm câu trả lời
1) 2sinx + cosx = sin2x + 1
2) (1 + cosx)(1+sinx) = 2
3) 3cos4x - 8cos6x + 2cos2x +3 =0
4) sin3x + cos3x.sinx + cosx = \(\sqrt{2}\)cos2x
5) (2cosx -1)(2sinx + cosx) = sin2x - sinx
Giải pt
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(sin2x-cos2x+3sinx-cosx-1=0\)
1.
\(2sin\left(x+\dfrac{\pi}{6}\right)+sinx+2cosx=3\)
\(\Leftrightarrow\sqrt{3}sinx+cosx+sinx+2cosx=3\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)sinx+3cosx=3\)
\(\Leftrightarrow\sqrt{13+2\sqrt{3}}\left[\dfrac{\sqrt{3}+1}{\sqrt{13+2\sqrt{3}}}sinx+\dfrac{3}{\sqrt{13+2\sqrt{3}}}cosx\right]=3\)
Đặt \(\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(pt\Leftrightarrow\sqrt{13+2\sqrt{3}}sin\left(x+\alpha\right)=3\)
\(\Leftrightarrow sin\left(x+\alpha\right)=\dfrac{3}{\sqrt{13+2\sqrt{3}}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\alpha=arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\\x+\alpha=\pi-arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm:
\(x=k2\pi;x=\pi-2arcsin\dfrac{3}{\sqrt{13+2\sqrt{3}}}+k2\pi\)
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2.
\(\left(sin2x+cos2x\right)cosx+2cos2x-sinx=0\)
\(\Leftrightarrow2sinx.cos^2x+cos2x.cosx+2cos2x-sinx=0\)
\(\Leftrightarrow\left(2cos^2x-1\right)sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.sinx+cos2x.cosx+2cos2x=0\)
\(\Leftrightarrow cos2x.\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow cos2x=0\)
\(\Leftrightarrow2x=\dfrac{\pi}{2}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
Vậy phương trình đã cho có nghiệm \(x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)
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