\(k\in Z\)
a.
\(cos\left(x-2\right)=\dfrac{2}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=arccos\left(\dfrac{2}{5}\right)+k2\pi\\x-2=-arccos\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+arccos\left(\dfrac{2}{5}\right)+k2\pi\\x=2-arcos\left(\dfrac{2}{5}\right)+k2\pi\end{matrix}\right.\)
d.
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{1}{2}\\cosx=3>1\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
f.
\(\Leftrightarrow cosx=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{3}+k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
h.
\(cos\left(3x+10^0\right)=-1\)
\(\Leftrightarrow3x+10^0=180^0+k360^0\)
\(\Leftrightarrow3x=170^0+k360^0\)
\(\Leftrightarrow x=\dfrac{1}{3}.170^0+k120^0\)
j.
\(cos\left[cos\left(x+2\right)\right]=1\)
\(\Leftrightarrow cos\left(x+2\right)=k2\pi\)
Do \(-1\le cos\left(x+2\right)\le1\Rightarrow-1\le k2\pi\le1\)
\(\Rightarrow k=0\)
\(\Rightarrow cos\left(x+2\right)=0\)
\(\Rightarrow x+2=\dfrac{\pi}{2}+n\pi\)
\(\Rightarrow x=-2+\dfrac{\pi}{2}+n\pi\)
Giúp em với ạ. Cần rất gấp ạ. Em cảm ơn ạ
