Chapter C, Problem 33E

To determine

To calculate:

The value of $x$ to satisfy the following inequality equation:

  $\sin x\le \frac{1}{2}$

Answer to Problem 33E

The values of $x$ for the interval $\left[0,2\pi \right]$ to satisfy eq. $\sin x\le \frac{1}{2}$ are $\left[0\le x\le \frac{\pi }{6}\right]$ and $\left[\frac{5\pi }{6}\le x\le \pi \right]$ .

Explanation of Solution

Given information:

  $\begin{array}{l}\sin x\le \frac{1}{2}\\ \left[0,2\pi \right]\end{array}$

Calculation:

Consider

The unit circle with co-ordinates $\left(x,y\right)=\left(\cos x,\sin x\right)$ and radius $1$ to find out the values of $x$ for the interval $\left[0,2\pi \right]$ :

  

Fig. Unit circle

From the unit circle figure:

The $y$ coordinate or sine value is equal to $\frac{1}{2}$ at $x=\frac{\pi }{6}$ and $x=\frac{5\pi }{6}$ .

Sine value is less than or equal to $\frac{1}{2}$ in the first quadrant for the $x$ value of interval $\left[0\le x\le \frac{\pi }{6}\right]$ . Sine is between $0$ and $\frac{1}{2}$ , for the $x$ value of interval $\left[\frac{5\pi }{6}\le x\le \pi \right]$ .

As in the entire fourth and third quadrant, the $y$ value is negative so, sine should be negative and it is less than $\frac{1}{2}$ for interval $\left[\pi \le x\le 2\pi \right]$ .

Therefore, the values of $x$ for the interval $\left[0,2\pi \right]$ to satisfy eq. $\sin x\le \frac{1}{2}$ are $\left[0\le x\le \frac{\pi }{6}\right]$ and $\left[\frac{5\pi }{6}\le x\le \pi \right]$ .