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\(cos3x=-cos\left(x-120^0\right)\)
\(\Leftrightarrow cos3x=cos\left(x+60^0\right)\)
\(\Rightarrow\left[{}\begin{matrix}3x=x+60^0+k360^0\\3x=-x-60^0+k360^0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=30^0+k180^0\\x=-15^0+k90^0\end{matrix}\right.\)
\(\Leftrightarrow sin\left(2x-90^0\right)=cos2x\)
\(\Leftrightarrow-cos2x=cos2x\)
\(\Rightarrow cos2x=0\Rightarrow2x=90^0+k180^0\)
\(\Rightarrow x=45^0+k90^0\)
\(cos^2x+sin^2x+2sinx.cosx=1+cos4x\)
\(\Leftrightarrow1+sin2x=1+cos4x\)
\(\Leftrightarrow cos4x=sin2x=cos\left(\frac{\pi}{2}-2x\right)\)
\(\Rightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}-2x+k2\pi\\4x=2x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{12}+\frac{k\pi}{3}\\x=-\frac{\pi}{4}+k\pi\end{matrix}\right.\)
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...
c/
\(\Leftrightarrow2sinx.cosx-2\sqrt{3}cos^2x=0\)
\(\Leftrightarrow2cosx\left(sinx-\sqrt{3}cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx-\sqrt{3}cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\sinx=\sqrt{3}cosx\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{sinx}{cosx}=\sqrt{3}\Leftrightarrow tanx=\sqrt{3}\)
\(\Rightarrow x=\frac{\pi}{3}+k\pi\)
d/
\(\Leftrightarrow tan\left(3x-50^0\right)=-cot\left(x-30^0\right)\)
\(\Leftrightarrow tan\left(3x-50^0\right)=tan\left(x+60^0\right)\)
\(\Rightarrow3x-50^0=x+60^0+k180^0\)
\(\Rightarrow x=55^0+k90^0\)
a/
\(\Leftrightarrow sinx=2cosx\)
Nhận thấy \(cosx=0\) không phải nghiệm, pt tương đương:
\(\frac{sinx}{cosx}=2\Leftrightarrow tanx=2\)
\(\Leftrightarrow tanx=tana\) (với \(a\in\left(0;\frac{\pi}{2}\right)\) sao cho \(tana=2\))
\(\Rightarrow x=a+k\pi\)
b/
\(tan2x=cotx=tan\left(\frac{\pi}{2}-x\right)\)
\(\Leftrightarrow2x=\frac{\pi}{2}-x+k\pi\)
\(\Rightarrow x=\frac{\pi}{6}+\frac{k\pi}{3}\)