let (f(x)=3.1 - 1.2secleft(\frac{pi x}{6}\right)). in the (xy) - plane, what are the (x) - coordinates of…

Explanation:

Step1: Set up the equation

Set $f(x)=-3$, so $3.1 - 1.2\sec(\frac{\pi x}{6})=-3$.

Step2: Isolate the secant - function

First, subtract 3.1 from both sides: $-1.2\sec(\frac{\pi x}{6})=-3 - 3.1=-6.1$. Then divide both sides by - 1.2: $\sec(\frac{\pi x}{6})=\frac{-6.1}{-1.2}=\frac{61}{12}$. Since $\sec\theta=\frac{1}{\cos\theta}$, we have $\cos(\frac{\pi x}{6})=\frac{12}{61}$.

Step3: Solve for x

We know that if $\cos\alpha = k$, then $\alpha = 2n\pi\pm\cos^{-1}(k)$, where $n\in\mathbb{Z}$. For $\cos(\frac{\pi x}{6})=\frac{12}{61}$, we have $\frac{\pi x}{6}=2n\pi\pm\cos^{-1}(\frac{12}{61})$. First, consider the principal - value case when $n = 0$. $\frac{\pi x}{6}=\cos^{-1}(\frac{12}{61})$ or $\frac{\pi x}{6}=2\pi-\cos^{-1}(\frac{12}{61})$. For $\frac{\pi x}{6}=\cos^{-1}(\frac{12}{61})$, then $x=\frac{6}{\pi}\cos^{-1}(\frac{12}{61})$. For $\frac{\pi x}{6}=2\pi-\cos^{-1}(\frac{12}{61})$, then $x = 12-\frac{6}{\pi}\cos^{-1}(\frac{12}{61})$. We are given the domain $0\leq x<2\pi\approx6.28$. $x=\frac{6}{\pi}\cos^{-1}(\frac{12}{61})\approx\frac{6}{\pi}\times1.41\approx2.69$ $x = 12-\frac{6}{\pi}\cos^{-1}(\frac{12}{61})\approx12 - 2.69 = 9.31$ (rejected as it is out of the domain $0\leq x<2\pi$)

Answer:

$\frac{6}{\pi}\cos^{-1}(\frac{12}{61})$