Đáp án:
$\begin{align}
& a)t=\dfrac{2}{9}+\dfrac{2k}{3};t=\dfrac{2k}{3} \\
& b)t=-\dfrac{1}{9}+\dfrac{2k}{3} \\
& c)t=0,76+\dfrac{2k}{3} \\
\end{align}$
Giải thích các bước giải:
a) thời điểm :
$\begin{align}
& x=10cos(3\pi t-\dfrac{\pi }{3}) \\
& \Leftrightarrow 5=10.cos(3\pi t-\dfrac{\pi }{3}) \\
& \Rightarrow 3\pi .t-\dfrac{\pi }{3}=\pm \dfrac{\pi }{3}+2k\pi \\
\end{align}$
$\Rightarrow \left\{ \begin{align}
& t=\dfrac{2}{9}s+\dfrac{2k}{3} \\
& t=\dfrac{2k}{3} \\
\end{align} \right.$
b) xét chiều:
$\begin{align}
& x=10cos(3\pi t-\dfrac{\pi }{3}) \\
& \Leftrightarrow 5=10.cos(3\pi t-\dfrac{\pi }{3}) \\
& \Rightarrow 3\pi .t-\dfrac{\pi }{3}=-\dfrac{2\pi }{3}+2k\pi \\
& \Rightarrow t=-\dfrac{1}{9}+\dfrac{2k}{3} \\
\end{align}$
c) chiều âm
$\begin{align}
& x=10cos(3\pi t-\dfrac{\pi }{3}) \\
& \Leftrightarrow \sqrt{5}=10.cos(3\pi t-\dfrac{\pi }{3}) \\
& \Rightarrow 3\pi t-\dfrac{\pi }{3}=0,43\pi +2k\pi \\
& \Rightarrow t=0,76+\dfrac{2k}{3} \\
\end{align}$
