10. let (f(x)=6.2 - 4.1csc(\frac{pi x}{4})). in the (xy - plane), what are the (x - coordinates) of the…

Explanation:

Step1: Set up the equation for problem 10

Set $f(x)=1$ in $f(x)=6.2 - 4.1\csc(\frac{\pi x}{4})$, so $1=6.2 - 4.1\csc(\frac{\pi x}{4})$.

Step2: Rearrange the equation

First, move the terms around: $4.1\csc(\frac{\pi x}{4})=6.2 - 1=5.2$. Then $\csc(\frac{\pi x}{4})=\frac{5.2}{4.1}=\frac{52}{41}$. Since $\csc\theta=\frac{1}{\sin\theta}$, we have $\sin(\frac{\pi x}{4})=\frac{41}{52}$.

Step3: Solve for $x$

$\frac{\pi x}{4}=\arcsin(\frac{41}{52})+2k\pi$ or $\frac{\pi x}{4}=\pi-\arcsin(\frac{41}{52})+2k\pi$, $k\in\mathbb{Z}$. For $0\leq x<\pi$, when $k = 0$, from $\frac{\pi x}{4}=\arcsin(\frac{41}{52})$, we get $x=\frac{4}{\pi}\arcsin(\frac{41}{52})$. From $\frac{\pi x}{4}=\pi-\arcsin(\frac{41}{52})$, we get $x = 4-\frac{4}{\pi}\arcsin(\frac{41}{52})$.

Step1: Set up the equation for problem 11

Set $f(x)= - 5$ in $f(x)=2.1+2.7\sec(3 - 0.4x)$, so $-5=2.1+2.7\sec(3 - 0.4x)$.

Step2: Rearrange the equation

Move the terms: $2.7\sec(3 - 0.4x)=-5 - 2.1=-7.1$. Then $\sec(3 - 0.4x)=-\frac{7.1}{2.7}=-\frac{71}{27}$. Since $\sec\theta=\frac{1}{\cos\theta}$, we have $\cos(3 - 0.4x)=-\frac{27}{71}$.

Step3: Solve for $x$

$3 - 0.4x=\arccos(-\frac{27}{71})+2k\pi$ or $3 - 0.4x = 2\pi-\arccos(-\frac{27}{71})+2k\pi$, $k\in\mathbb{Z}$. Then $0.4x=3-\arccos(-\frac{27}{71})-2k\pi$ or $0.4x=3-(2\pi-\arccos(-\frac{27}{71}))-2k\pi$. For $0\leq x<2\pi$, when $k = 0$, $x=\frac{3-\arccos(-\frac{27}{71})}{0.4}$ or $x=\frac{3-(2\pi-\arccos(-\frac{27}{71}))}{0.4}$.

Answer:

For problem 10: $x=\frac{4}{\pi}\arcsin(\frac{41}{52}),4-\frac{4}{\pi}\arcsin(\frac{41}{52})$ For problem 11: $x=\frac{3-\arccos(-\frac{27}{71})}{0.4},\frac{3-(2\pi-\arccos(-\frac{27}{71}))}{0.4}$