giải phương trình
a) \(cosx=3\)
b) \(cosx=\dfrac{\sqrt{3}}{2}\)
c) \(cos\left(x+\dfrac{\pi}{12}\right)\) = \(cos\dfrac{3\pi}{12}\)
d) \(cos\)(x - 20 độ) = cos70 độ
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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: \(2x< >k\Omega\)
=>\(x< >\dfrac{k\Omega}{2}\)
\(cot2x=-\dfrac{\sqrt{3}}{3}\)
=>\(cot2x=cot\left(-\dfrac{\Omega}{3}\right)\)
=>\(2x=-\dfrac{\Omega}{3}+k\Omega\)
=>\(x=-\dfrac{\Omega}{6}+\dfrac{k\Omega}{2}\)
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)
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
NL
Giải phương trình: \(\sqrt{3}cosx+\dfrac{3}{cos^2\left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)}=6+sinx\)
0
22 tháng 8 2023
a: cos5x=-5
mà -1<=cos5x<=1
nên \(x\in\varnothing\)
b: 2*cosx-1=0
=>2*cosx=1
=>cosx=1/2
=>x=pi/3+k2pi hoặc x=-pi/3+k2pi
c: -5*cos(x+pi/3)=0
=>cos(x+pi/3)=0
=>x+pi/3=pi/2+kpi
=>x=pi/6+kpi
d: cos4x=cos(5/12pi)
=>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi
=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2
e: cos^2x=1
=>sin^2x=0
=>sin x=0
=>x=kpi
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 \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
6 tháng 11 2023
a: \(sinx=sin\left(\dfrac{\Omega}{4}\right)\)
=>\(\left[{}\begin{matrix}x=\dfrac{\Omega}{4}+k2\Omega\\x=\Omega-\dfrac{\Omega}{4}+k2\Omega=\dfrac{3}{4}\Omega+k2\Omega\end{matrix}\right.\)
b: cos2x=cosx
=>\(\left[{}\begin{matrix}2x=x+k2\Omega\\2x=-x+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2\Omega\\3x=k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=k2\Omega\\x=\dfrac{k2\Omega}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{k2\Omega}{3}\)
c:
ĐKXĐ: \(x-\dfrac{\Omega}{3}< >\dfrac{\Omega}{2}+k\Omega\)
=>\(x< >\dfrac{5}{6}\Omega+k\Omega\)
\(tan\left(x-\dfrac{\Omega}{3}\right)=\sqrt{3}\)
=>\(x-\dfrac{\Omega}{3}=\dfrac{\Omega}{3}+k\Omega\)
=>\(x=\dfrac{2}{3}\Omega+k\Omega\)
d:
ĐKXĐ: \(2x+\dfrac{\Omega}{6}< >k\Omega\)
=>\(2x< >-\dfrac{\Omega}{6}+k\Omega\)
=>\(x< >-\dfrac{1}{12}\Omega+\dfrac{k\Omega}{2}\)
\(cot\left(2x+\dfrac{\Omega}{6}\right)=cot\left(\dfrac{\Omega}{4}\right)\)
=>\(2x+\dfrac{\Omega}{6}=\dfrac{\Omega}{4}+k\Omega\)
=>\(2x=\dfrac{1}{12}\Omega+k\Omega\)
=>\(x=\dfrac{1}{24}\Omega+\dfrac{k\Omega}{2}\)
a: cosx=3
mà -1<=cosx<=1
nên \(x\in\varnothing\)
b: \(cosx=\dfrac{\sqrt{3}}{2}\)
=>\(cosx=cos\left(\dfrac{pi}{6}\right)\)
=>x=pi/6+k2pi hoặc x=-pi/6+k2pi
c: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{pi}{12}=\dfrac{3}{12}pi+k2pi\\x+\dfrac{pi}{12}=-\dfrac{3}{12}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{12}pi+k2pi=\dfrac{1}{6}pi+k2pi\\x=-\dfrac{4}{12}pi+k2pi=-\dfrac{1}{3}pi+k2pi\end{matrix}\right.\)
d: =>x-20 độ=70 độ+k*360 hoặc x-20=-70+k*360
=>x=90+k*360 hoặc x=-50+k*360