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? 
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.
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.
| $$\left(0, \frac{π}{6}\right)$$ | $$\left(\frac{π}{6}, \frac{5π}{6}\right)$$ | $$\left(\frac{5π}{6}, 2π\right)$$ |
|---|---|---|
| (-) | (+) | (-) |
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.
| $$\left(0, \frac{2π}{3}\right)$$ | $$\left(\frac{2π}{3}, π\right)$$ | $$\left(π, \frac{4π}{3}\right)$$ | $$\left(\frac{4π}{3}, 2π\right)$$ |
|---|
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.
| $$\left(0, \frac{2π}{3}\right)$$ | $$\left(\frac{2π}{3}, π\right)$$ | $$\left(π, \frac{4π}{3}\right)$$ | $$\left(\frac{4π}{3}, 2π\right)$$ |
|---|---|---|---|
| (+) | (-) | (-) | (+) |
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.
| $$\left(0, \frac{π}{6}\right)$$ | $$\left(\frac{π}{6}, \frac{π}{2}\right)$$ | $$\left(\frac{π}{2}, \frac{5π}{6}\right)$$ | $$\left(\frac{5π}{6}, π\right)$$ | $$\left(π, 2π\right)$$ |
|---|
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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