Chapter 0, Problem 58RE

a).

To determine

To solve: the given equation in the given interval.

Answer to Problem 58RE

The solutions are $x=\pi +0.201,x=2\pi -0.201$ .

Explanation of Solution

Given:

  $\sin x=-0.2,0\le x<2\pi$ .

Calculation:

Consider,

  $\begin{array}{c}\sin x=0.2\\ x={\sin }^{-1}\left(0.2\right)\\ x=0.201\end{array}$

Here $x=0.201$ is in the first quadrant.

Now the angle $\pi +0.201$ is in the third quadrant and has sine equals $-0.2$ .

And the angle $2\pi -0.201$ is in the fourth quadrant and has sine equals $-0.2$ .

Thus, the solutions are

  $x=\pi +0.201,x=2\pi -0.201$

b).

To determine

To solve: the given equation in the given interval.

Answer to Problem 58RE

The solutions are $x=\pi +0.201+2k\pi$ , where $k$ is any integer.

Explanation of Solution

Given:

  $\sin x=-0.2,0\le x<2\pi$ .

Calculation:

From subpart (a), one solution of the equation $\sin x=-0.2$ is $x=\pi +0.201$ .

Since $\sin x$ is periodic with period $2\pi$ , the solutions to the given equation are of the form

  $x=\pi +0.201+2k\pi$ , where $k$ is any integer.