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a: \(cos\left(2x-\dfrac{\Omega}{6}\right)+cos\left(x+\dfrac{\Omega}{3}\right)=0\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)+sin\left(\dfrac{\Omega}{6}-x\right)=0\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)=-sin\left(\dfrac{\Omega}{6}-x\right)=sin\left(x-\dfrac{\Omega}{6}\right)\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)=cos\left(\dfrac{\Omega}{2}-x+\dfrac{\Omega}{6}\right)\)
=>\(cos\left(2x-\dfrac{\Omega}{6}\right)=cos\left(-x+\dfrac{2}{3}\Omega\right)\)
=>\(\left[{}\begin{matrix}2x-\dfrac{\Omega}{6}=-x+\dfrac{2\Omega}{3}+k2\Omega\\2x-\dfrac{\Omega}{6}=x-\dfrac{2}{3}\Omega+k2\Omega\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=\dfrac{5}{6}\Omega+k2\Omega\\x=-\dfrac{1}{2}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{18}\Omega+\dfrac{k2\Omega}{3}\\x=-\dfrac{1}{2}\Omega+k2\Omega\end{matrix}\right.\)
b: \(cos\left(2x+30^0\right)+sin\left(x-30^0\right)=0\)
=>\(cos\left(2x+30^0\right)=-sin\left(x-30^0\right)\)
=>\(cos\left(2x+30^0\right)=sin\left(-x+30^0\right)\)
=>\(cos\left(2x+30^0\right)=cos\left(60^0+x\right)\)
=>\(\left[{}\begin{matrix}2x+30^0=x+60^0+k\cdot360^0\\2x+30^0=-x-60^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=30^0+k\cdot360^0\\3x=-90^0+k\cdot360^0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=30^0+k\cdot360^0\\x=-30^0+k\cdot120^0\end{matrix}\right.\)
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 ...
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...
a: =>2sin(x+pi/3)=-1
=>sin(x+pi/3)=-1/2
=>x+pi/3=-pi/6+k2pi hoặc x+pi/3=7/6pi+k2pi
=>x=-1/2pi+k2pi hoặc x=2/3pi+k2pi
b: =>2sin(x-30 độ)=-1
=>sin(x-30 độ)=-1/2
=>x-30 độ=-30 độ+k*360 độ hoặc x-30 độ=180 độ+30 độ+k*360 độ
=>x=k*360 độ hoặc x=240 độ+k*360 độ
c: =>2sin(x-pi/6)=-căn 3
=>sin(x-pi/6)=-căn 3/2
=>x-pi/6=-pi/3+k2pi hoặc x-pi/6=4/3pi+k2pi
=>x=-1/6pi+k2pi hoặc x=3/2pi+k2pi
d: =>2sin(x+10 độ)=-căn 3
=>sin(x+10 độ)=-căn 3/2
=>x+10 độ=-60 độ+k*360 độ hoặc x+10 độ=240 độ+k*360 độ
=>x=-70 độ+k*360 độ hoặc x=230 độ+k*360 độ
e: \(\Leftrightarrow2\cdot sin\left(x-15^0\right)=-\sqrt{2}\)
=>\(sin\left(x-15^0\right)=-\dfrac{\sqrt{2}}{2}\)
=>x-15 độ=-45 độ+k*360 độ hoặc x-15 độ=225 độ+k*360 độ
=>x=-30 độ+k*360 độ hoặc x=240 độ+k*360 độ
f: \(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=-\dfrac{1}{\sqrt{2}}\)
=>x-pi/3=-pi/4+k2pi hoặc x-pi/3=5/4pi+k2pi
=>x=pi/12+k2pi hoặc x=19/12pi+k2pi
g) \(3+\sqrt[]{5}sin\left(x+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=-\dfrac{3}{\sqrt[]{5}}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right)=sin\left[arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)\right]\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\\x+\dfrac{\pi}{3}=\pi-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}-arcsin\left(-\dfrac{3}{\sqrt[]{5}}\right)+k2\pi\end{matrix}\right.\)
h) \(1+sin\left(x-30^o\right)=0\)
\(\Leftrightarrow sin\left(x-30^o\right)=-1\)
\(\Leftrightarrow sin\left(x-30^o\right)=sin\left(-90^o\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-30^o=-90^0+k360^o\\x-30^o=180^o+90^0+k360^o\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-60^0+k360^o\\x=300^0+k360^o\end{matrix}\right.\)
\(\Leftrightarrow x=-60^0+k360^o\)
a: \(2\cdot cot\left(\dfrac{pi}{2}-x\right)+tan\left(pi-x\right)\)
\(=2\cdot tanx-tanx\)
=tan x
b: \(sin\left(\dfrac{5}{2}pi-x\right)+cos\left(13pi+x\right)-sin\left(x-5pi\right)\)
\(=sin\left(\dfrac{pi}{2}-x\right)+cos\left(pi+x\right)+sin\left(pi-x\right)\)
\(=cosx-cosx+sinx=sinx\)
1.
ĐKXĐ: \(x\ne k\pi\)
\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
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
\(cos3x=-cos\left(x-120^0\right)\)
\(\Leftrightarrow cos3x=cos\left(x+60^0\right)\)
\(\Rightarrow\left[{}\begin{matrix}3x=x+60^0+k360^0\\3x=-x-60^0+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-15^0+k90^0\end{matrix}\right.\)
\(\Leftrightarrow sin\left(2x-90^0\right)=cos2x\)
\(\Leftrightarrow-cos2x=cos2x\)
\(\Rightarrow cos2x=0\Rightarrow2x=90^0+k180^0\)
\(\Rightarrow x=45^0+k90^0\)
\(cos^2x+sin^2x+2sinx.cosx=1+cos4x\)
\(\Leftrightarrow1+sin2x=1+cos4x\)
\(\Leftrightarrow cos4x=sin2x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}-2x+k2\pi\\4x=2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{3}\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)