Giải pt
\(32\sin^6\dfrac{x}{2}+\sin3x=3\sin x\)
Những câu hỏi liên quan
a) \(sinx=-\dfrac{6}{5}\)
b) \(sin3x=\dfrac{\sqrt{3}}{2}\)
c) \(sin\left(x+\dfrac{\pi}{3}\right)=sin\dfrac{3\pi}{4}\)
d) \(4sin\left(x+\dfrac{5\pi}{6}\right)=5\)
a: sin x=-6/5=-1,2
mà -1<=sin x<=1
nên \(x\in\varnothing\)
b: sin3x=căn 3/2
=>3x=pi/3+k2pi hoặc 3x=2/3pi+k2pi
=>x=pi/9+k2pi/3 hoặc x=2/9pi+k2pi/3
c: \(sin\left(x+\dfrac{pi}{3}\right)=sin\left(\dfrac{3}{4}pi\right)\)
=>x+pi/3=3/4pi+k2pi hoặc x+pi/3=1/4pi+k2pi
=>x=5/12pi+k2pi hoặc x=-1/12pi+k2pi
d: =>sin(x+5/6pi)=5/4
mà sin(x+5/6pi) thuộc [-1;1]
nên \(x\in\varnothing\)
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Giải các PT sau
1. \(\cos^2\left(x-30^{\cdot}\right)-\sin^2\left(x-30^{\cdot}\right)=\sin\left(x+60^{\cdot}\right)\)
2. \(\sin^22x+\cos^23x=1\)
3. \(\sin x+\sin2x+\sin3x+\sin4x=0\)
4. \(\sin^2x+\sin^22x=\sin^23x\)
1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
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giải pt
a, \(\sin^2x+\sin^22x+\sin^23x=\dfrac{3}{2}\)
b. \(\cos^2x+\sin^22x+\cos^23x=1\)
c,\(\sin5x+2\cos^2x=1\)
d,\(1+\tan x=2\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\)
e,\(\sin3x+\cos3x-\sin x+\cos x=\sqrt{2}\cos2x\)
Giai các pt sau
1. \(\sqrt{3}\cos5x-2\sin3x.\cos2x-\sin x=0\)
4. \(\sin3x+\cos3x-\sin x+\cos x=\sqrt{2}\cos2x\)
6. \(\sin x+\cos x.\sin2x+\sqrt{3}\cos3x=2\left(\cos4x+\sin x^3\right)\)
a/sin3x+cos2x1+2sin xcos2xb/sin^3x+cos^3x2left(sin^5x+cos^5xright)c/dfrac{tan x}{sin x}-dfrac{sin x}{cos x}dfrac{sqrt{2}}{2}d/dfrac{cos xleft(cos x+2sin xright)+3sin xleft(sin x+sqrt{2}right)}{sin2x-1}1e/sin^2x+sin^23x-2cos^22x0f/dfrac{tan x-sin x}{sin^3x}dfrac{1}{cos x}g/sin2xleft(cos x+tan2xright)4cos^2xh/sin^2x+sin^23xcos^2x+cos^23xk/4sin2xdfrac{cos^2x-sin^2x}{cos^6x+sin^6x}mọi người giải giúp em với em đang cần gấp ạ
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a/\(\sin3x+\cos2x=1+2\sin x\cos2x\)
b/\(\sin^3x+\cos^3x=2\left(\sin^5x+\cos^5x\right)\)
c/\(\dfrac{\tan x}{\sin x}-\dfrac{\sin x}{\cos x}=\dfrac{\sqrt{2}}{2}\)
d/\(\dfrac{\cos x\left(\cos x+2\sin x\right)+3\sin x\left(\sin x+\sqrt{2}\right)}{\sin2x-1}=1\)
e/\(\sin^2x+\sin^23x-2\cos^22x=0\)
f/\(\dfrac{\tan x-\sin x}{\sin^3x}=\dfrac{1}{\cos x}\)
g/\(\sin2x\left(\cos x+\tan2x\right)=4\cos^2x\)
h/\(\sin^2x+\sin^23x=\cos^2x+\cos^23x\)
k/\(4\sin2x=\dfrac{\cos^2x-\sin^2x}{\cos^6x+\sin^6x}\)
mọi người giải giúp em với em đang cần gấp ạ
Giải PTa1) dfrac{left(1-2sin xright)cos x}{left(1+2sin xright)left(1-sin xright)}sqrt{3}a2) 2sin17x+sqrt{3}cos5x+sin5x0a3) cos7x-sin5xsqrt{3}left(cos5x-sin7xright)a4) sqrt{3}cos5x-2sin3xcos2x-sin x0a5) tan x+cot x2left(sin2x+cos2xright)
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Giải PT
a1) \(\dfrac{\left(1-2\sin x\right)\cos x}{\left(1+2\sin x\right)\left(1-\sin x\right)}=\sqrt{3}\)
a2) \(2\sin17x+\sqrt{3}\cos5x+\sin5x=0\)
a3) \(\)\(\cos7x-\sin5x=\sqrt{3}\left(\cos5x-\sin7x\right)\)
a4) \(\sqrt{3}\cos5x-2\sin3x\cos2x-\sin x=0\)
a5) \(\tan x+\cot x=2\left(\sin2x+\cos2x\right)\)
a) \(\sin2x=\dfrac{-1}{2}\)
b) \(\sin\dfrac{x}{2}=-1\)
c) \(\sin3x=\dfrac{1}{3}\)
d) \(\sin4x=-\sin x\)
\(a,\sin2x=\dfrac{-1}{2}\Leftrightarrow\sin2x=\sin\left(-\dfrac{\pi}{6}\right)\\ \Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{6}+k2\pi\\2x=\pi+\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k2\pi\\x=\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\left(k\in Z\right)\)
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a, \(sin2x=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{6}+k2\pi\\2x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{7\pi}{12}+k\pi\end{matrix}\right.\)
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b, \(sin\dfrac{x}{2}=-1\)
\(\Leftrightarrow\dfrac{x}{2}=-\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\pi+k2\pi\)
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Giải các PT sau:1. dfrac{left(2cos2x-1right)left(sin x-3right)}{sin x}0 2.dfrac{3left(sin x+cos xright)}{sin x-cos x}2+2cos x3.dfrac{3left(sin x+tan xright)}{tan x-sin x}-2cos x24. 1+sin x+cos x+sin2x+cos2x05. 2sin xleft(1+cos2xright)+sin2x1+2cos x
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Giải các PT sau:
1. \(\dfrac{\left(2\cos2x-1\right)\left(\sin x-3\right)}{\sin x}=0\)
2.\(\dfrac{3\left(\sin x+\cos x\right)}{\sin x-\cos x}=2+2\cos x\)
3.\(\dfrac{3\left(\sin x+\tan x\right)}{\tan x-\sin x}-2\cos x=2\)
4. \(1+\sin x+\cos x+\sin2x+\cos2x=0\)
5. \(2\sin x\left(1+\cos2x\right)+\sin2x=1+2\cos x\)
1.
ĐKXĐ: \(x\ne k\pi\)
\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
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2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.
3.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)
\(\Leftrightarrow\dfrac{3\left(1+cosx\right)}{1-cosx}+2\left(1+cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(\dfrac{3}{1-cosx}+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\left(loại\right)\\cosx=\dfrac{5}{2}\left(loại\right)\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
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4.
\(\Leftrightarrow\left(sin^2x+cos^2x+2sinx.cosx\right)+\left(sinx+cosx\right)+\left(cos^2x-sin^2x\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)^2+\left(sinx+cosx\right)+\left(sinx+cosx\right)\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(sinx+cosx+1+cosx-sinx\right)=0\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(2cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
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22. Tìm nghiệm dương nhỏ nhất của PT: \(3\sin^2x+2\sin x\cos x-\cos^2x=0\)
23. Giải PT: \(\sqrt{3}\cos x+2\sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{1}\right)=1\)
\(\sqrt{3}cosx+2sin^2\left(\dfrac{x}{2}-\pi\right)=1\)
\(\Leftrightarrow\sqrt{3}cosx+2sin^2\dfrac{x}{2}=1\)
\(\Leftrightarrow\sqrt{3}cosx-cosx=0\Leftrightarrow cosx=0\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) ( k thuộc Z )
Vậy ...
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22.
Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)
\(3tan^2x+2tanx-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{3}\right)+k\pi\end{matrix}\right.\)
Nghiệm dương nhỏ nhất của pt là: \(x=arctan\left(\dfrac{1}{3}\right)\)
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22. PT đã cho tương đương
3 - 4cos2x + 2 sinxcosx = 0
⇔ 3 - 2 - 2cos2x + sin2x = 0
⇔ 1 - 2cos2x + sin2x = 0
⇔ 1 + sin2x = 2cos2x
⇔ sin\(\dfrac{\pi}{2}\) + sin2x = 2cos2x
⇔ \(2sin\left(\dfrac{\pi}{4}+x\right).cos\left(\dfrac{\pi}{4}-x\right)\) = 2cos2x
Do \(\left(\dfrac{\pi}{4}-x\right)+\left(\dfrac{\pi}{4}+x\right)=\dfrac{\pi}{2}\)
⇒ \(sin\left(\dfrac{\pi}{4}+x\right)=cos\left(\dfrac{\pi}{4}-x\right)\)
Vậy sin2\(\left(x+\dfrac{\pi}{4}\right)\) = cos2x
Cái này là hiển nhiên ????
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