Lesson Objectives

- Demonstrate the ability to solve a trigonometric equation
- Learn how to solve a trigonometric inequality

## How to Solve a Trigonometric Inequality

Over the course of the last few lessons, we have learned how to solve most types of trigonometric equations. How can we solve a trigonometric inequality?

Example #1: Solve each inequality for 0 ≤ x < 2$π$. $$4 - \text{sin}\hspace{.1em}x > \frac{7}{2}$$ 1) Identify the domain of the trigonometric function in the problem:

The sine function is defined for all real numbers, so we can move on to step 2.

2) Use the addition property of inequality to change the right side into 0 and then simplify the left side of the inequality as much as possible: $$4 - \text{sin}\hspace{.1em}x > \frac{7}{2}$$ Subtract 7/2 away from each side of the inequality: $$4 - \text{sin}\hspace{.1em}x - \frac{7}{2}> 0$$ Simplify the left side: $$- \text{sin}\hspace{.1em}x + \frac{8}{2}- \frac{7}{2}> 0$$ $$- \text{sin}\hspace{.1em}x + \frac{1}{2}> 0$$ Multiply both sides by -1, recall this requires that we flip the direction of the inequality symbol: $$\text{sin}\hspace{.1em}x - \frac{1}{2}< 0$$ 3) Find the critical values: $$\text{sin}\hspace{.1em}x - \frac{1}{2}=0$$ Add 1/2 to both sides: $$\text{sin}\hspace{.1em}x=\frac{1}{2}$$ In our given interval, x = $π$/6, 5$π$/6.

4) Split the number line up into intervals based on the critical values:

Here, you have a lot of room to use what you want. You can set up a traditional number line, use a table, or even use the unit circle. Here, we will use a basic table since it's easiest to read.

Note: We have excluded 0, $π$/6, $5π$/6, and 2$π$. These values will be considered at the end.

5) Test a point in each interval.

Here, we just need to pick a value in each interval and plug into our simplified inequality: $$\text{sin}\hspace{.1em}x - \frac{1}{2}< 0$$ Alternatively, we can just think about the sign.

6) We can use our table to put a solution together, note that $π$/6 and 5$π$/6 are not included since this is a strict inequality. Additionally, 0 is included and 2$π$ is excluded because of the interval we are solving over. $$0≤ x < \frac{π}{6}$$ $$\text{or}$$ $$\frac{5π}{6}< x < 2π$$ Interval Notation: $$\left[0, \frac{π}{6}\right) ∪ \left(\frac{5π}{6}, 2π\right)$$ We can also show this graphically: Example #2: Solve each inequality for 0 ≤ x < 2$π$. $$-2\text{cos}^{2}x ≥ 1 + 3\text{cos}\hspace{.1em}x$$ 1) Identify the domain of the trigonometric function in the problem:

The cosine function is defined for all real numbers, so we can move on to step 2.

2) Use the addition property of inequality to change the right side into 0 and then simplify the left side of the inequality as much as possible: $$-2\text{cos}^{2}x ≥ 1 + 3\text{cos}\hspace{.1em}x$$ Subtract 3cos x away from each side of the inequality: $$-2\text{cos}^{2}x - 3\text{cos}\hspace{.1em}x≥ 1$$ Subtract 1 away from each side of the inequality: $$-2\text{cos}^{2}x - 3\text{cos}\hspace{.1em}x - 1≥ 0$$ Multiply both sides by -1, recall this requires that we flip the direction of the inequality symbol: $$2\text{cos}^{2}x + 3\text{cos}\hspace{.1em}x + 1 ≤ 0$$ 3) Find the critical values: $$2\text{cos}^{2}x + 3\text{cos}\hspace{.1em}x + 1=0$$ Factor the left side: $$(2\text{cos}\hspace{.1em}x + 1)(\text{cos}\hspace{.1em}x + 1)=0$$ Use the zero-product property: $$2\text{cos}\hspace{.1em}x + 1=0$$ $$\text{or}$$ $$\text{cos}\hspace{.1em}x + 1=0$$ Top Equation: $$2\text{cos}\hspace{.1em}x + 1=0$$ Subtract 1 away from each side and then divide both sides by 2: $$\text{cos}\hspace{.1em}x=-\frac{1}{2}$$ $$x=\frac{2π}{3}, \frac{4π}{3}$$ Bottom Equation: $$\text{cos}\hspace{.1em}x + 1=0$$ Subtract 1 away from each side: $$\text{cos}\hspace{.1em}x=-1$$ $$x=π$$ In our given interval: $$x=\frac{2π}{3}, π, \frac{4π}{3}$$ 4) Split the number line up into intervals based on the critical values:

Here, you have a lot of room to use what you want. You can set up a traditional number line, use a table, or even use the unit circle. Here, we will use a basic table since it's easiest to read.

5) Test a point in each interval.

Here, we just need to pick a value in each interval and plug into our simplified inequality: $$(2\text{cos}\hspace{.1em}x + 1)(\text{cos}\hspace{.1em}x + 1) ≤ 0$$ Alternatively, we can just think about the sign.

6) We can use our table to put a solution together, note that $2π$/3, $π$, and 4$π$/3 are included since this is a non-strict inequality. $$\frac{2π}{3}≤ x ≤ \frac{4π}{3}$$ Interval Notation: $$\left[\frac{2π}{3}, \frac{4π}{3}\right]$$ We can also show this graphically: Example #3: Solve each inequality for 0 ≤ x < 2$π$. $$\text{cot}^2x-3\text{csc}\hspace{.1em}x+3> 0$$ 1) Identify the domain of the trigonometric function in the problem:

In our interval, the cotangent function is undefined at 0 and $π$, additionally, the cosecant function will be undefined at 0 and $π$ as well.

2) Use the addition property of inequality to change the right side into 0 and then simplify the left side of the inequality as much as possible:

Here, the right side is already zero. We will use the following identity to simplify: $$\text{cot}^2x=\text{csc}^2x - 1$$ Plug in for $\text{cot}^2x$: $$\text{cot}^2x-3\text{csc}\hspace{.1em}x+3> 0$$ $$\text{csc}^2x - 1-3\text{csc}\hspace{.1em}x+3> 0$$ Simplify: $$\text{csc}^2x - 3\text{csc}\hspace{.1em}x + 2> 0$$ 3) Find the critical values: $$\text{csc}^2x - 3\text{csc}\hspace{.1em}x + 2=0$$ Factor the left side: $$(\text{csc}\hspace{.1em}x - 2)(\text{csc}\hspace{.1em}x - 1)=0$$ Use the zero-product property: $$\text{csc}\hspace{.1em}x - 2=0$$ $$\text{or}$$ $$\text{csc}\hspace{.1em}x - 1=0$$ Top Equation: $$\text{csc}\hspace{.1em}x - 2=0$$ Add 2 to each side: $$\text{csc}\hspace{.1em}x=2$$ $$x=\frac{π}{6}, \frac{5π}{6}$$ Bottom Equation: $$\text{csc}\hspace{.1em}x - 1=0$$ Add 1 to each side: $$\text{csc}\hspace{.1em}x=1$$ $$x=\frac{π}{2}$$ In our given interval: $$x=\frac{π}{6}, \frac{π}{2}, \frac{5π}{6}$$ 4) Split the number line up into intervals based on the critical values:

Here, you have a lot of room to use what you want. You can set up a traditional number line, use a table, or even use the unit circle. Here, we will use a basic table since it's easiest to read.

Note: In this problem, we have 0, and $π$ that are restricted from our domain in the given interval. These need to be included when we set up our table.

5) Test a point in each interval.

Here, we just need to pick a value in each interval and plug into our simplified inequality: $$(\text{csc}\hspace{.1em}x - 2)(\text{csc}\hspace{.1em}x - 1) > 0$$ Alternatively, we can just think about the sign.

6) We can use our table to put a solution together, note that our critical values are all excluded since this is a strict inequality. $$0 < x < \frac{π}{6}$$ $$\text{or}$$ $$\frac{5π}{6}< x < π$$ $$\text{or}$$ $$π < x < 2π$$ Interval Notation: $$\left(0, \frac{π}{6}\right) ∪ \left(\frac{5π}{6}, π\right) ∪ \left(π, 2π\right)$$ We can also show this graphically:

### General Method for Solving a Trigonometric Inequality

- Identify the domain of the trigonometric function in the problem
- Remember that some trigonometric functions have restricted domains (e.g., the tangent function is undefined at odd multiples of $π$/2)

- Use the addition property of inequality to change the right side into 0 and then simplify the left side of the inequality as much as possible
- Find the critical values
- Replace the inequality symbol with an equality symbol and solve the resulting equation
- These critical values are where the function changes sign
- Critical values also occur where the function is undefined, found in step 1

- Split the number line up into intervals based on the critical values
- Solution(s) to our related equation
- Any undefined points

- Test a point in each interval
- Use a sign chart to keep track of the sign of the function
- Consider the critical values separately

- The solution is the union of all intervals where the function is positive or negative, depending on the inequality sign given in the problem

### Unit Circle

The unit circle will be given here for reference: Let's begin with a very simple example.Example #1: Solve each inequality for 0 ≤ x < 2$π$. $$4 - \text{sin}\hspace{.1em}x > \frac{7}{2}$$ 1) Identify the domain of the trigonometric function in the problem:

The sine function is defined for all real numbers, so we can move on to step 2.

2) Use the addition property of inequality to change the right side into 0 and then simplify the left side of the inequality as much as possible: $$4 - \text{sin}\hspace{.1em}x > \frac{7}{2}$$ Subtract 7/2 away from each side of the inequality: $$4 - \text{sin}\hspace{.1em}x - \frac{7}{2}> 0$$ Simplify the left side: $$- \text{sin}\hspace{.1em}x + \frac{8}{2}- \frac{7}{2}> 0$$ $$- \text{sin}\hspace{.1em}x + \frac{1}{2}> 0$$ Multiply both sides by -1, recall this requires that we flip the direction of the inequality symbol: $$\text{sin}\hspace{.1em}x - \frac{1}{2}< 0$$ 3) Find the critical values: $$\text{sin}\hspace{.1em}x - \frac{1}{2}=0$$ Add 1/2 to both sides: $$\text{sin}\hspace{.1em}x=\frac{1}{2}$$ In our given interval, x = $π$/6, 5$π$/6.

4) Split the number line up into intervals based on the critical values:

Here, you have a lot of room to use what you want. You can set up a traditional number line, use a table, or even use the unit circle. Here, we will use a basic table since it's easiest to read.

$$\left(0, \frac{π}{6}\right)$$ | $$\left(\frac{π}{6}, \frac{5π}{6}\right)$$ | $$\left(\frac{5π}{6}, 2π\right)$$ |
---|

5) Test a point in each interval.

Here, we just need to pick a value in each interval and plug into our simplified inequality: $$\text{sin}\hspace{.1em}x - \frac{1}{2}< 0$$ Alternatively, we can just think about the sign.

$$\left(0, \frac{π}{6}\right)$$ | $$\left(\frac{π}{6}, \frac{5π}{6}\right)$$ | $$\left(\frac{5π}{6}, 2π\right)$$ |
---|---|---|

(-) | (+) | (-) |

The cosine function is defined for all real numbers, so we can move on to step 2.

2) Use the addition property of inequality to change the right side into 0 and then simplify the left side of the inequality as much as possible: $$-2\text{cos}^{2}x ≥ 1 + 3\text{cos}\hspace{.1em}x$$ Subtract 3cos x away from each side of the inequality: $$-2\text{cos}^{2}x - 3\text{cos}\hspace{.1em}x≥ 1$$ Subtract 1 away from each side of the inequality: $$-2\text{cos}^{2}x - 3\text{cos}\hspace{.1em}x - 1≥ 0$$ Multiply both sides by -1, recall this requires that we flip the direction of the inequality symbol: $$2\text{cos}^{2}x + 3\text{cos}\hspace{.1em}x + 1 ≤ 0$$ 3) Find the critical values: $$2\text{cos}^{2}x + 3\text{cos}\hspace{.1em}x + 1=0$$ Factor the left side: $$(2\text{cos}\hspace{.1em}x + 1)(\text{cos}\hspace{.1em}x + 1)=0$$ Use the zero-product property: $$2\text{cos}\hspace{.1em}x + 1=0$$ $$\text{or}$$ $$\text{cos}\hspace{.1em}x + 1=0$$ Top Equation: $$2\text{cos}\hspace{.1em}x + 1=0$$ Subtract 1 away from each side and then divide both sides by 2: $$\text{cos}\hspace{.1em}x=-\frac{1}{2}$$ $$x=\frac{2π}{3}, \frac{4π}{3}$$ Bottom Equation: $$\text{cos}\hspace{.1em}x + 1=0$$ Subtract 1 away from each side: $$\text{cos}\hspace{.1em}x=-1$$ $$x=π$$ In our given interval: $$x=\frac{2π}{3}, π, \frac{4π}{3}$$ 4) Split the number line up into intervals based on the critical values:

Here, you have a lot of room to use what you want. You can set up a traditional number line, use a table, or even use the unit circle. Here, we will use a basic table since it's easiest to read.

$$\left(0, \frac{2π}{3}\right)$$ | $$\left(\frac{2π}{3}, π\right)$$ | $$\left(π, \frac{4π}{3}\right)$$ | $$\left(\frac{4π}{3}, 2π\right)$$ |
---|

Here, we just need to pick a value in each interval and plug into our simplified inequality: $$(2\text{cos}\hspace{.1em}x + 1)(\text{cos}\hspace{.1em}x + 1) ≤ 0$$ Alternatively, we can just think about the sign.

$$\left(0, \frac{2π}{3}\right)$$ | $$\left(\frac{2π}{3}, π\right)$$ | $$\left(π, \frac{4π}{3}\right)$$ | $$\left(\frac{4π}{3}, 2π\right)$$ |
---|---|---|---|

(+) | (-) | (-) | (+) |

In our interval, the cotangent function is undefined at 0 and $π$, additionally, the cosecant function will be undefined at 0 and $π$ as well.

2) Use the addition property of inequality to change the right side into 0 and then simplify the left side of the inequality as much as possible:

Here, the right side is already zero. We will use the following identity to simplify: $$\text{cot}^2x=\text{csc}^2x - 1$$ Plug in for $\text{cot}^2x$: $$\text{cot}^2x-3\text{csc}\hspace{.1em}x+3> 0$$ $$\text{csc}^2x - 1-3\text{csc}\hspace{.1em}x+3> 0$$ Simplify: $$\text{csc}^2x - 3\text{csc}\hspace{.1em}x + 2> 0$$ 3) Find the critical values: $$\text{csc}^2x - 3\text{csc}\hspace{.1em}x + 2=0$$ Factor the left side: $$(\text{csc}\hspace{.1em}x - 2)(\text{csc}\hspace{.1em}x - 1)=0$$ Use the zero-product property: $$\text{csc}\hspace{.1em}x - 2=0$$ $$\text{or}$$ $$\text{csc}\hspace{.1em}x - 1=0$$ Top Equation: $$\text{csc}\hspace{.1em}x - 2=0$$ Add 2 to each side: $$\text{csc}\hspace{.1em}x=2$$ $$x=\frac{π}{6}, \frac{5π}{6}$$ Bottom Equation: $$\text{csc}\hspace{.1em}x - 1=0$$ Add 1 to each side: $$\text{csc}\hspace{.1em}x=1$$ $$x=\frac{π}{2}$$ In our given interval: $$x=\frac{π}{6}, \frac{π}{2}, \frac{5π}{6}$$ 4) Split the number line up into intervals based on the critical values:

Here, you have a lot of room to use what you want. You can set up a traditional number line, use a table, or even use the unit circle. Here, we will use a basic table since it's easiest to read.

Note: In this problem, we have 0, and $π$ that are restricted from our domain in the given interval. These need to be included when we set up our table.

$$\left(0, \frac{π}{6}\right)$$ | $$\left(\frac{π}{6}, \frac{π}{2}\right)$$ | $$\left(\frac{π}{2}, \frac{5π}{6}\right)$$ | $$\left(\frac{5π}{6}, π\right)$$ | $$\left(π, 2π\right)$$ |
---|

Here, we just need to pick a value in each interval and plug into our simplified inequality: $$(\text{csc}\hspace{.1em}x - 2)(\text{csc}\hspace{.1em}x - 1) > 0$$ Alternatively, we can just think about the sign.

$$\left(0, \frac{π}{6}\right)$$ | $$\left(\frac{π}{6}, \frac{π}{2}\right)$$ | $$\left(\frac{π}{2}, \frac{5π}{6}\right)$$ | $$\left(\frac{5π}{6}, π\right)$$ | $$\left(π, 2π\right)$$ |
---|---|---|---|---|

(+) | (-) | (-) | (+) | (+) |

#### Skills Check:

Example #1

Solve each inequality for 0 ≤ x < 2$π$. $$\text{cos}\hspace{.1em}x + 3 < \frac{6 - \sqrt{3}}{2}$$

Please choose the best answer.

A

$$\frac{5π}{6}< x < \frac{7π}{6}$$

B

$$\frac{5π}{6}≤ x ≤ \frac{7π}{6}$$

C

$$\frac{π}{6}< x < \frac{5π}{6}$$

D

$$\frac{3π}{4}< x < \frac{7π}{4}$$

E

$$\frac{π}{2}< x < π$$

Example #2

Solve each inequality for 0 ≤ x < 2$π$. $$\text{cos}\left(x + \frac{5π}{4}\right) + 4 ≥ \frac{7}{2}$$ Hint: Make a substitution.

Please choose the best answer.

A

$$\frac{π}{6}≤ x ≤ \frac{17π}{9}$$

B

$$\frac{π}{3}≤ x ≤ \frac{5π}{3}$$

C

$$\frac{7π}{4}≤ x ≤ \frac{11π}{6}$$

D

$$\frac{π}{12}≤ x ≤ \frac{17π}{12}$$

E

$$\text{No Solution}$$

Example #3

Solve each inequality for 0 ≤ x < 2$π$. $$2\text{sin}^{2}x + 3\text{sin}\hspace{.1em}x + 1 ≤ 0$$

A

$$\frac{π}{12}≤ x ≤ \frac{π}{6}$$

B

$$\frac{7π}{6}≤ x ≤ \frac{11π}{6}$$

C

$$\text{No Solution}$$

D

$$\text{All Real Numbers}$$

E

$$\frac{π}{3}≤ x ≤ \frac{2π}{3}$$

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