## Trigonometry (11th Edition) Clone

$$t=\frac{1}{6}$$
$$V=\cos2\pi t$$ where $t\in[0,\frac{1}{2}]$ As $V=0.5$, we would have the equation: $$\cos2\pi t=0.5=\frac{1}{2}$$ We would take the interval $[0,2\pi)$ here. Over the interval $[0,2\pi)$, there are 2 values of $2\pi t$ where $\cos2\pi t=\frac{1}{2}$, which are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$. That means $$2\pi t=\frac{\pi}{3}\hspace{1cm}\text{or}\hspace{1cm}2\pi t=\frac{5\pi}{3}$$ $$t=\frac{\pi}{3\times2\pi}\hspace{1cm}\text{or}\hspace{1cm}t=\frac{5\pi}{3\times2\pi}$$ $$t=\frac{1}{6}\hspace{1cm}\text{or}\hspace{1cm}t=\frac{5}{6}$$ $t=\frac{5}{6}$ lies out of the defined interval for $t$: $t\in[0,\frac{1}{2}]$, so we need to eliminate $t=\frac{5}{6}$ That means $t=\frac{1}{6}$ is the result of this question.