Chapter 5, Problem 15PS

a.

To determine

  $\sin x\ge 0.5$

Answer to Problem 15PS

  $\boxed{x=\left[\frac{\pi }{6},\frac{5\pi }{6}\right]}$

Explanation of Solution

Given information:

Find the solution of each inequality in the interval $[0,2\pi )$ .

Calculation:

The inequality involving the function, can be solved as follows:

  $\sin x\ge 0.5$

  $\sin x-\frac{1}{2}\ge 0$

The values of $x$ for which the right hand side of the above equation remains non-negative are the solutions to given inequality. Since $\sin x$ is greater than or equal to $\frac{1}{2}$ for the interval $\left[\frac{\pi }{6},\frac{5\pi }{6}\right]$ . So, the solution for $x$ is,

  $x=\left[\frac{\pi }{6},\frac{5\pi }{6}\right]$

b.

To determine

  $\cos x\le -0.5$

Answer to Problem 15PS

  $\boxed{x=\left[\frac{2\pi }{3},\frac{4\pi }{3}\right]}$

Explanation of Solution

Given information:

Find the solution of each inequality in the interval $[0,2\pi )$ .

Calculation:

The inequality involving the function, can be solved as follows:

  $\cos x\le -0.5$

  $\cos x-\frac{1}{2}\le 0$

The values of $x$ for which the right hand side of the above equation remains non-negative are the solutions to given inequality. Since $\sin x$ is greater than or equal to $\frac{1}{2}$ for the interval $\left[\frac{2\pi }{3},\frac{4\pi }{3}\right]$ . So, the solution for $x$ is,

  $\boxed{x=\left[\frac{2\pi }{3},\frac{4\pi }{3}\right]}$

c.

To determine

  $\tan x<\sin x$

Answer to Problem 15PS

  $\boxed{x=\left(\frac{\pi }{2},\pi \right)}$

Explanation of Solution

Given information:

Find the solution of each inequality in the interval $[0,2\pi )$ .

Calculation:

The inequality involving the function, can be solved as follows:

  $\tan x<\sin x$

  $\begin{array}{l}\frac{\sin x}{\cos x}<\sin x\\ \sin x\left(\frac{1}{\cos x}-1\right)<0\\ \sin x\left(\frac{1-\cos x}{\cos x}\right)<0\end{array}$

If the $\cos x$ in the denominator of the above expression is negative, then it might change the sign of the inequality. So, we need to multiply both the sides by ${\cos }^{2}x$ of the inequality to prevent the change of sign. We can do this as follows:

  $\begin{array}{l}\sin x\left(\frac{1-\cos x}{\cos x}\right){\cos }^{2}x<0\times {\cos }^{2}x\\ \sin x\cos x\left(1-\cos x\right)<0\end{array}$

Since $1-\cos x$ will always be positive as the minimum value of $1-\cos x$ is non-negative. So, deleting $1-\cos x$ from the numerator will not affect the result.

  $\begin{array}{l}\sin x\cos x<0\\ 2\sin x\cos x<0\times {\cos }^{2}x\\ \sin 2x<0\end{array}$

Now, the values of $x$ for which $\sin x$ is positive will be:

  $\begin{array}{l}2x=\left(\pi ,2\pi \right)\\ x=\left(\frac{\pi }{2},\pi \right)\end{array}$

Hence, the values of $x$ for which $\sin x$ is positive is $\boxed{x=\left(\frac{\pi }{2},\pi \right)}$ .

d.

To determine

  $\cos x\ge \sin x$

Answer to Problem 15PS

  $\boxed{x=\left(\frac{\pi }{2},\pi \right)}$

Explanation of Solution

Given information:

Find the solution of each inequality in the interval $[0,2\pi )$ .

Calculation:

The inequality involving the function, can be solved as follows:

  $\cos x\ge \sin x$

  $\cos x-\sin x\ge 0$

Now we multiply the above inequality by in order to further simplify the inequality as follows:

  $\begin{array}{l}\frac{1}{\sqrt{2}}\left(\sin x-\cos x\right)\ge 0\\ \frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x\ge 0\\ \sin \left(x-{\cos }^{-1}\frac{1}{\sqrt{2}}\right)\ge 0\\ \sin \left(x-\frac{\pi }{4}\right)\ge 0\end{array}$

Now, the values of $x$ for which $\sin \left(x-\frac{\pi }{4}\right)$ is positive will be:

  $\begin{array}{l}x-\frac{\pi }{4}\ge \left(0,\pi \right)\\ x\ge \left(\frac{\pi }{4},\frac{5\pi }{4}\right)\end{array}$

Hence, the values of $x$ for which $\sin \left(x-\frac{\pi }{4}\right)$ is positive is $\boxed{x=\left(\frac{\pi }{2},\pi \right)}$ .