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Pt \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{3}-x=2x+\dfrac{\pi}{3}+k2\pi\\\dfrac{\pi}{3}-x=-2x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-k2\pi}{3}\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...
\(cos\left(x-\dfrac{\pi}{3}\right)=sin\left(2x+\dfrac{\pi}{2}\right)\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=cos2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x-\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{\pi}{3}-x+l2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{9}+l\dfrac{2\pi}{3}\end{matrix}\right.\)
Chỉ II đúng
<=> 2cos (3x/2)cos(x/2+pi/3)=0
<=>cos (3x/2)=0 hoặc cos (x/2+pi/3)=0
<=>3x/2=pi/2+kpi hoặc x/2+pi/3=pi/2+kpi (k thuộc z)
<=>x=pi/3+(2/3)kpi hoặc x=pi/3+2kpi (k thuộc z)
KL
\(cos\left(2x+\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow2cos\dfrac{3x}{2}.cos\left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\dfrac{3x}{2}=0\\cos\left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=\dfrac{\pi}{2}+k\pi\\\dfrac{x}{2}+\dfrac{\pi}{3}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)