Giải các phương trình :
a) \(\cos\left(x+3\right)=\dfrac{1}{3}\)
b) \(\cos\left(3x-45^0\right)=\dfrac{\sqrt{3}}{2}\)
c) \(\cos\left(2x+\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\)
d) \(\left(2+\cos x\right)\left(3\cos2x-1\right)=0\)
K
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Những câu hỏi liên quan
30 tháng 7 2023
d: cos^2x=1
=>sin^2x=0
=>sin x=0
=>x=kpi
a: =>sin 4x=cos(x+pi/6)
=>sin 4x=sin(pi/2-x-pi/6)
=>sin 4x=sin(pi/3-x)
=>4x=pi/3-x+k2pi hoặc 4x=2/3pi+x+k2pi
=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3
b: =>x+pi/3=pi/6+k2pi hoặc x+pi/3=-pi/6+k2pi
=>x=-pi/2+k2pi hoặc x=-pi/6+k2pi
c: =>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi
=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2023
a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2023
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
22 tháng 8 2023
a: cos3x=8
mà -1<=cos3x<=1
nên \(x\in\varnothing\)
b; \(-2\cdot cosx+\sqrt{3}=0\)
=>\(-2\cdot cosx=-\sqrt{3}\)
=>\(cosx=\dfrac{\sqrt{3}}{2}\)
=>x=pi/6+k2pi hoặc x=-pi/6+k2pi
c: cos(3x-pi/6)=0
=>3x-pi/6=pi/2+k2pi
=>3x=2/3pi+k2pi
=>x=2/9pi+k2pi/3
d: cos(x+2/3pi)=cos(pi/5)
=>x+2/3pi=pi/5+k2pi hoặc x+2/3pi=-pi/5+k2pi
=>x=-7/15pi+k2pi hoặc x=-13/15pi+k2pi
e: cos^2(3x)=4
=>cos3x=2(loại) hoặc cos3x=-2(loại)
5 tháng 9 2023
1: cos(2x+pi/6)=cos(pi/3-3x)
=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=3x-pi/3+k2pi
=>5x=pi/6+k2pi hoặc -x=-1/2pi+k2pi
=>x=pi/30+k2pi/5 hoặc x=pi-k2pi
2: sin(2x+pi/6)=sin(pi/3-3x)
=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=pi-pi/3+3x+k2pi
=>5x=pi/6+k2pi hoặc -x=2/3pi-pi/6+k2pi
=>x=pi/30+k2pi/5 hoặc x=-1/2pi-k2pi
ND
Nguyễn Đức Trí
VIP
6 tháng 9 2023
1) \(cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{3}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-3x+k2\pi\\2x+\dfrac{\pi}{6}=-\dfrac{\pi}{3}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{3}-\dfrac{\pi}{6}+k2\pi\\3x-2x=\dfrac{\pi}{3}+\dfrac{\pi}{6}-k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\) \(\left(k\in N\right)\)
23 tháng 6 2021
a, Ta có : \(\sin\left(3x+60\right)=\dfrac{1}{2}\)
\(\Rightarrow3x+60=30+2k180\)
\(\Rightarrow3x=2k180-30\)
\(\Leftrightarrow x=120k-10\)
Vậy ...
b, Ta có : \(\cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Rightarrow2x-\dfrac{\pi}{3}=\dfrac{3}{4}\pi+k2\pi\)
\(\Leftrightarrow x=\dfrac{13}{24}\pi+k\pi\)
Vậy ...
c, Ta có : \(tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)
\(\Rightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)
Vậy ...
d, Ta có : \(\cot\left(2x+\pi\right)=-1\)
\(\Rightarrow2x+\pi=\dfrac{3}{4}\pi+k\pi\)
\(\Leftrightarrow x=-\dfrac{1}{8}\pi+\dfrac{k}{2}\pi\)
Vậy ...
23 tháng 6 2021
a) \(sin\left(3x+60^0\right)=\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(3x+\dfrac{\pi}{3}\right)=sin\dfrac{\pi}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))
Vậy...
b) Pt\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\dfrac{3\pi}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13\pi}{24}+k\pi\\x=-\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)(\(k\in Z\))
Vậy...
c) Pt \(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=tan\dfrac{\pi}{3}\)
\(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi,k\in Z\)\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi,k\in Z\)
Vậy...
d) Pt \(\Leftrightarrow tan\left(2x+\pi\right)=-1\)
\(\Leftrightarrow2x+\pi=-\dfrac{\pi}{4}+k\pi,k\in Z\)
\(\Leftrightarrow x=-\dfrac{5\pi}{8}+\dfrac{k\pi}{2},k\in Z\)
Vậy...
6 tháng 11 2023
a: \(sin\left(x-\dfrac{\Omega}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
=>\(sin\left(x-\dfrac{\Omega}{4}\right)=sin\left(-\dfrac{\Omega}{4}\right)\)
=>\(\left[{}\begin{matrix}x-\dfrac{\Omega}{4}=-\dfrac{\Omega}{4}+k2\Omega\\x-\dfrac{\Omega}{4}=\Omega+\dfrac{\Omega}{4}+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{3}{2}\Omega+k2\Omega\end{matrix}\right.\)
b: \(cos\left(x+\dfrac{\Omega}{4}\right)=cos\left(\dfrac{3}{4}\Omega\right)\)
=>\(\left[{}\begin{matrix}x+\dfrac{\Omega}{4}=\dfrac{3}{4}\Omega+k2\Omega\\x+\dfrac{\Omega}{4}=-\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\Omega+k2\Omega\\x=-\Omega+k2\Omega\end{matrix}\right.\)
c: ĐKXĐ: \(\left\{{}\begin{matrix}2x< >\dfrac{\Omega}{2}+k\Omega\\x+\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{4}+\dfrac{k\Omega}{2}\\x< >\dfrac{1}{6}\Omega+k\Omega\end{matrix}\right.\)
\(tan2x=tan\left(x+\dfrac{\Omega}{3}\right)\)
=>\(2x=x+\dfrac{\Omega}{3}+k\Omega\)
=>\(x=\dfrac{\Omega}{3}+k\Omega\)
d: ĐKXĐ: \(2x< >k\Omega\)
=>\(x< >\dfrac{k\Omega}{2}\)
\(cot2x=-\dfrac{\sqrt{3}}{3}\)
=>\(cot2x=cot\left(-\dfrac{\Omega}{3}\right)\)
=>\(2x=-\dfrac{\Omega}{3}+k\Omega\)
=>\(x=-\dfrac{\Omega}{6}+\dfrac{k\Omega}{2}\)
5 tháng 9 2021
1, \(\left(sinx+\dfrac{sin3x+cos3x}{1+2sin2x}\right)=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+2sinx.sin2x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{sinx+cosx+sin3x}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{2sin2x.cosx+cosx}{1+2sin2x}=\dfrac{3+cos2x}{5}\)
⇔ \(\dfrac{cosx\left(2sin2x+1\right)}{1+2sin2x}=\dfrac{2+2cos^2x}{5}\)
⇒ cosx = \(\dfrac{2+2cos^2x}{5}\)
⇔ 2cos2x - 5cosx + 2 = 0
⇔ \(\left[{}\begin{matrix}cosx=2\\cosx=\dfrac{1}{2}\end{matrix}\right.\)
⇔ \(x=\pm\dfrac{\pi}{3}+k.2\pi\) , k là số nguyên
2, \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\left(1+cot2x.cotx\right)=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cos2x.cosx+sin2x.sinx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2}{sin^2x}.\dfrac{cosx}{sin2x.sinx}=0\)
⇔ \(48-\dfrac{1}{cos^4x}-\dfrac{2cosx}{2cosx.sin^4x}=0\)
⇒ \(48-\dfrac{1}{cos^4x}-\dfrac{1}{sin^4x}=0\). ĐKXĐ : sin2x ≠ 0
⇔ \(\dfrac{1}{cos^4x}+\dfrac{1}{sin^4x}=48\)
⇒ sin4x + cos4x = 48.sin4x . cos4x
⇔ (sin2x + cos2x)2 - 2sin2x. cos2x = 3 . (2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}\) . (2sinx . cosx)2 = 3(2sinx.cosx)4
⇔ 1 - \(\dfrac{1}{2}sin^22x\) = 3sin42x
⇔ \(sin^22x=\dfrac{1}{2}\) (thỏa mãn ĐKXĐ)
⇔ 1 - 2sin22x = 0
⇔ cos4x = 0
⇔ \(x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)
5 tháng 9 2021
3, \(sin^4x+cos^4x+sin\left(3x-\dfrac{\pi}{4}\right).cos\left(x-\dfrac{\pi}{4}\right)-\dfrac{3}{2}=0\)
⇔ \(\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x+\dfrac{1}{2}sin\left(4x-\dfrac{\pi}{2}\right)+\dfrac{1}{2}sin2x-\dfrac{3}{2}=0\)
⇔ \(1-\dfrac{1}{2}sin^22x+\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{3}{2}=0\)
⇔ \(\dfrac{1}{2}sin2x-\dfrac{1}{2}cos4x-\dfrac{1}{2}-\dfrac{1}{2}sin^22x=0\)
⇔ sin2x - sin22x - (1 + cos4x) = 0
⇔ sin2x - sin22x - 2cos22x = 0
⇔ sin2x - 2 (cos22x + sin22x) + sin22x = 0
⇔ sin22x + sin2x - 2 = 0
⇔ \(\left[{}\begin{matrix}sin2x=1\\sin2x=-2\end{matrix}\right.\)
⇔ sin2x = 1
⇔ \(2x=\dfrac{\pi}{2}+k.2\pi\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
4, cos5x + cos2x + 2sin3x . sin2x = 0
⇔ cos5x + cos2x + cosx - cos5x = 0
⇔ cos2x + cosx = 0
⇔ \(2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)
⇔ \(cos\dfrac{3x}{2}=0\)
⇔ \(\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\)
⇔ x = \(\dfrac{\pi}{3}+k.\dfrac{2\pi}{3}\)
Do x ∈ [0 ; 2π] nên ta có \(0\le\dfrac{\pi}{3}+k\dfrac{2\pi}{3}\le2\pi\)
⇔ \(-\dfrac{1}{2}\le k\le\dfrac{5}{2}\). Do k là số nguyên nên k ∈ {0 ; 1 ; 2}
Vậy các nghiệm thỏa mãn là các phần tử của tập hợp
\(S=\left\{\dfrac{\pi}{3};\pi;\dfrac{5\pi}{3}\right\}\)
TL
1 tháng 12 2019
Chứng minh các biểu thức đã cho không phụ thuộc vào x.
Từ đó suy ra f'(x)=0
a) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
b) f(x)=1⇒f′(x)=0f(x)=1⇒f′(x)=0 ;
c) f(x)=\(\frac{1}{4}\)(\(\sqrt{2}\)-\(\sqrt{6}\))=>f'(x)=0
d,f(x)=\(\frac{3}{2}\)=>f'(x)=0
