Chapter C, Problem 34E

To determine

To calculate:

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

  $2\cos x+1>0$

Answer to Problem 34E

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:

  $\begin{array}{l}2\cos x+1>0\\ \left[0,2\pi \right]\end{array}$

Calculation:

Know that:

  $\begin{array}{l}2\cos x+1>0\\ 2\cos x>-1\\ \cos x>-\frac{1}{2}\end{array}$

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 $\left[0,2\pi \right]$ that cause $\cos x>-\frac{1}{2}$ .

  

Fig. Unit circle

From the unit circle figure:

The $x$ coordinate or cosine value is positive and should greater than $-\frac{1}{2}$ , is observed in the first and fourth quadrant.

So, the values gets as $\left[0\le x\le \frac{\pi }{2}\right]$ and $\left[\frac{3\pi }{2}\le x\le 2\pi \right]$ .

Cosine value is greater than or equal to $-\frac{1}{2}$ in the third and second quadrant for the $x$ value of interval $\left[0\le x\le \frac{\pi }{6}\right]$ .

In the first quadrant $\cos x>-\frac{1}{2}$ , $x$ value of interval $\left[\frac{\pi }{2}\le x\le \frac{2\pi }{3}\right]$

In the third quadrant $\cos x>-\frac{1}{2}$ , $x$ value of interval $\left[\frac{4\pi }{3}\le x\le \frac{3\pi }{2}\right]$

Therefore, the values of $x$ for the interval $\left[0,2\pi \right]$ to satisfy eq. $2\cos x+1>0$ are $\left[0\le x\le \frac{2\pi }{3}\right]$ and $\left[\frac{4\pi }{3}\le x\le 2\pi \right]$ .