solve sin(x)=0.2 on 0≤x<2π. there are two solutions, a and b, with a < b. a = .201 b = give your answers…

solve sin(x)=0.2 on 0≤x<2π. there are two solutions, a and b, with a < b. a = .201 b = give your answers accurate to 3 decimal places. question help: video message instructor submit question

solve sin(x)=0.2 on 0≤x<2π. there are two solutions, a and b, with a < b. a = .201 b = give your answers accurate to 3 decimal places. question help: video message instructor submit question

Answer

Explanation:

Step1: Recall inverse - sine property

If $\sin(x)=a$, then $x = \sin^{-1}(a)+2k\pi$ or $x=\pi-\sin^{-1}(a)+2k\pi$, $k\in\mathbb{Z}$. Given $\sin(x) = 0.2$ and $0\leq x<2\pi$. First, $x_1=\sin^{-1}(0.2)$. $x_1=\sin^{-1}(0.2)\approx0.201$ (in radians).

Step2: Find the second solution

The second solution in the interval $[0, 2\pi)$ is $x_2=\pi-\sin^{-1}(0.2)$. $x_2=\pi - 0.201\approx3.142 - 0.201=2.941$.

Answer:

$B = 2.941$