Lesson Objectives

- Learn how to solve basic trigonometric equations
- Learn how to solve trigonometric equations using linear methods

## How to Solve Trigonometric Equations Using Linear Methods

In the last section of our course, we studied trigonometric identities. Recall that identities are equations that are true for all values of the variables for which all expressions are defined. Now, we will learn how to work with trigonometric equations that are conditional. Conditional equations are satisfied by some values but not others.

Example #1: Solve each equation over the interval [0°, 360°) and for all solutions. $$2 + 4\text{sin}\hspace{.1em}θ=4$$ Our goal is to isolate sin θ.

Subtract 2 away from each side: $$4\text{sin}\hspace{.1em}θ=2$$ Divide both sides by 4: $$\text{sin}\hspace{.1em}θ=\frac{1}{2}$$ We know that sine is positive in quadrants I and II only. If we reference our unit circle above, we obtain two values for θ in our interval of 150° and 30°.

Notice that the reference angle for 150° is 30°. If you don't have a unit circle handy, we can always use our inverse sine function on a calculator. This will only give us one solution, we must then use the concept of a reference angle to find the other. $$\text{sin}^{-1}\left(\frac{1}{2}\right)=30°$$ Since sine is positive in quadrants I and II, we can find the angle in quadrant II with a reference angle of 30°. This will be our other solution.

QII Reference Angle: $$θ'=180° - θ$$ $$θ=180° - θ'$$ $$180° - 30°=150°$$ This tells us that a 150° angle has a reference angle of 30°. Therefore, sine of 150° will be the same as sine of 30°. Our two solutions for θ are 30° and 150°. $$\{30°, 150°\}$$ or using radians: $$\left\{\frac{π}{6}, \frac{5π}{6}\right\}$$ To find all solutions, recall that the period of sine and cosine is 360° or $2π$ radians and the period of tangent is 180° or $π$ radians. Here, we will just add integer multiples of the period of the sine function. $$\{30° + 360°n, 150° + 360°n\}$$ $$\left\{\frac{π}{6}+ 2πn, \frac{5π}{6}+ 2πn\right\}$$ where n is any integer. Let's look at another example.

Example #2: Solve each equation over the interval [0°, 360°) and for all solutions. $$-4 - 4\text{cos}\hspace{.1em}θ=-3 - 2\text{cos}\hspace{.1em}θ$$ Our goal is to isolate cos θ.

Add 4 to each side: $$- 4\text{cos}\hspace{.1em}θ=1 - 2\text{cos}\hspace{.1em}θ$$ Add $2\text{cos}\hspace{.1em}θ$ to both sides: $$- 2\text{cos}\hspace{.1em}θ=1$$ Divide both sides by -2: $$\text{cos}\hspace{.1em}θ=-\frac{1}{2}$$ We know that cosine is negative in quadrants II and III only. If we reference our unit circle above, we obtain two values for θ in our interval of 120° and 240°.

Notice that the reference angle for 120° is 60°. Again, we can use our calculator to find the inverse cosine function with -1/2 as the argument. $$\text{cos}^{-1}\left(-\frac{1}{2}\right)=120°$$ Since cosine is negative in quadrants II and III, we can find the angle in quadrant III with a reference angle of 60°. This will be our other solution.

QIII Reference Angle: $$θ'=θ - 180°$$ $$θ=180° + θ'$$ $$180° + 60°=240°$$ Our two solutions for θ are 120° and 240°. $$\{120°, 240°\}$$ or using radians: $$\left\{\frac{2π}{3}, \frac{4π}{3}\right\}$$ With cosine the period is also 360° or $2π$ in terms of radians. Again, to find all solutions, we will just add integer multiples of the period of the cosine function. $$\{120° + 360°n, 240° + 360°n\}$$ or using radians: $$\left\{\frac{2π}{3}+ 2πn, \frac{4π}{3}+ 2πn\right\}$$ where n is any integer. Let's look at another example.

Example #3: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$-3\sqrt{3}+ 4 - \text{tan}\hspace{.1em}θ=4 + 8\text{tan}\hspace{.1em}θ$$ Our goal is to isolate tan θ.

Subtract 4 away from each side: $$-3\sqrt{3}- \text{tan}\hspace{.1em}θ=8\text{tan}\hspace{.1em}θ$$ Subtract 8 tan θ away from each side: $$-3\sqrt{3}- 9\text{tan}\hspace{.1em}θ=0$$ Add $3\sqrt{3}$ to each side: $$- 9\text{tan}\hspace{.1em}θ=3\sqrt{3}$$ Divide each side by -9: $$\text{tan}\hspace{.1em}θ=-\frac{\sqrt{3}}{3}$$ In this case, it is probably easiest to use the inverse tangent function on our calculator. $$\text{tan}^{-1}\left(-\frac{\sqrt{3}}{3}\right)=-30°$$ Since we are thinking about the interval [0°, 360°), we will add 360° to obtain 330°. This is not our only solution. Recall that tangent is negative in quadrants II and IV. We want to find an angle in quadrant II with a 30° reference angle. $$180° - 30°=150°$$ Our two solutions for θ are 150° and 330°. $$\{150°, 330°\}$$ or using radians: $$\left\{\frac{5π}{6}, \frac{11π}{6}\right\}$$ To find all solutions, we add integer multiples of the period of the tangent function, which is $π$. $$\{150° + 180°n, 330° + 180°n\}$$ You may realize that 150° + 180°(1) = 330°.

This allows us to write a more compact solution: $$\{150° + 180°n\}$$ or using radians: $$\left\{\frac{5π}{6}+ πn\right\}$$

### Unit Circle

The unit circle will be given here for reference.### Solving Trigonometric Equations Using Linear Methods

Let's begin with a very simple type of trigonometric equation. These can be solved using our basic properties of equalities to isolate a trigonometric expression on one side of the equation. Let's look at a few examples.Example #1: Solve each equation over the interval [0°, 360°) and for all solutions. $$2 + 4\text{sin}\hspace{.1em}θ=4$$ Our goal is to isolate sin θ.

Subtract 2 away from each side: $$4\text{sin}\hspace{.1em}θ=2$$ Divide both sides by 4: $$\text{sin}\hspace{.1em}θ=\frac{1}{2}$$ We know that sine is positive in quadrants I and II only. If we reference our unit circle above, we obtain two values for θ in our interval of 150° and 30°.

Notice that the reference angle for 150° is 30°. If you don't have a unit circle handy, we can always use our inverse sine function on a calculator. This will only give us one solution, we must then use the concept of a reference angle to find the other. $$\text{sin}^{-1}\left(\frac{1}{2}\right)=30°$$ Since sine is positive in quadrants I and II, we can find the angle in quadrant II with a reference angle of 30°. This will be our other solution.

QII Reference Angle: $$θ'=180° - θ$$ $$θ=180° - θ'$$ $$180° - 30°=150°$$ This tells us that a 150° angle has a reference angle of 30°. Therefore, sine of 150° will be the same as sine of 30°. Our two solutions for θ are 30° and 150°. $$\{30°, 150°\}$$ or using radians: $$\left\{\frac{π}{6}, \frac{5π}{6}\right\}$$ To find all solutions, recall that the period of sine and cosine is 360° or $2π$ radians and the period of tangent is 180° or $π$ radians. Here, we will just add integer multiples of the period of the sine function. $$\{30° + 360°n, 150° + 360°n\}$$ $$\left\{\frac{π}{6}+ 2πn, \frac{5π}{6}+ 2πn\right\}$$ where n is any integer. Let's look at another example.

Example #2: Solve each equation over the interval [0°, 360°) and for all solutions. $$-4 - 4\text{cos}\hspace{.1em}θ=-3 - 2\text{cos}\hspace{.1em}θ$$ Our goal is to isolate cos θ.

Add 4 to each side: $$- 4\text{cos}\hspace{.1em}θ=1 - 2\text{cos}\hspace{.1em}θ$$ Add $2\text{cos}\hspace{.1em}θ$ to both sides: $$- 2\text{cos}\hspace{.1em}θ=1$$ Divide both sides by -2: $$\text{cos}\hspace{.1em}θ=-\frac{1}{2}$$ We know that cosine is negative in quadrants II and III only. If we reference our unit circle above, we obtain two values for θ in our interval of 120° and 240°.

Notice that the reference angle for 120° is 60°. Again, we can use our calculator to find the inverse cosine function with -1/2 as the argument. $$\text{cos}^{-1}\left(-\frac{1}{2}\right)=120°$$ Since cosine is negative in quadrants II and III, we can find the angle in quadrant III with a reference angle of 60°. This will be our other solution.

QIII Reference Angle: $$θ'=θ - 180°$$ $$θ=180° + θ'$$ $$180° + 60°=240°$$ Our two solutions for θ are 120° and 240°. $$\{120°, 240°\}$$ or using radians: $$\left\{\frac{2π}{3}, \frac{4π}{3}\right\}$$ With cosine the period is also 360° or $2π$ in terms of radians. Again, to find all solutions, we will just add integer multiples of the period of the cosine function. $$\{120° + 360°n, 240° + 360°n\}$$ or using radians: $$\left\{\frac{2π}{3}+ 2πn, \frac{4π}{3}+ 2πn\right\}$$ where n is any integer. Let's look at another example.

Example #3: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$-3\sqrt{3}+ 4 - \text{tan}\hspace{.1em}θ=4 + 8\text{tan}\hspace{.1em}θ$$ Our goal is to isolate tan θ.

Subtract 4 away from each side: $$-3\sqrt{3}- \text{tan}\hspace{.1em}θ=8\text{tan}\hspace{.1em}θ$$ Subtract 8 tan θ away from each side: $$-3\sqrt{3}- 9\text{tan}\hspace{.1em}θ=0$$ Add $3\sqrt{3}$ to each side: $$- 9\text{tan}\hspace{.1em}θ=3\sqrt{3}$$ Divide each side by -9: $$\text{tan}\hspace{.1em}θ=-\frac{\sqrt{3}}{3}$$ In this case, it is probably easiest to use the inverse tangent function on our calculator. $$\text{tan}^{-1}\left(-\frac{\sqrt{3}}{3}\right)=-30°$$ Since we are thinking about the interval [0°, 360°), we will add 360° to obtain 330°. This is not our only solution. Recall that tangent is negative in quadrants II and IV. We want to find an angle in quadrant II with a 30° reference angle. $$180° - 30°=150°$$ Our two solutions for θ are 150° and 330°. $$\{150°, 330°\}$$ or using radians: $$\left\{\frac{5π}{6}, \frac{11π}{6}\right\}$$ To find all solutions, we add integer multiples of the period of the tangent function, which is $π$. $$\{150° + 180°n, 330° + 180°n\}$$ You may realize that 150° + 180°(1) = 330°.

This allows us to write a more compact solution: $$\{150° + 180°n\}$$ or using radians: $$\left\{\frac{5π}{6}+ πn\right\}$$

#### Skills Check:

Example #1

Solve for 0 ≤ θ < $2π$ $$-3 - 3\text{cot}\hspace{.1em}θ=-2 - 4\text{cot}\hspace{.1em}θ$$

Please choose the best answer.

A

$$\left\{\frac{π}{2}, \frac{3π}{2}, \frac{5π}{6}\right\}$$

B

$$\left\{\frac{π}{2}\right\}$$

C

$$\left\{\frac{π}{4}, \frac{5π}{4}\right\}$$

D

$$\left\{\frac{5π}{3}, \frac{7π}{6}\right\}$$

E

$$\left\{\frac{π}{12}, \frac{2π}{3}, \frac{11π}{6}\right\}$$

Example #2

Solve for 0 ≤ θ < $2π$ $$-3-\frac{1}{2}\text{tan}\hspace{.1em}θ=\frac{-6 - \sqrt{3}}{2}- 2\text{tan}\hspace{.1em}θ$$

Please choose the best answer.

A

$$\left\{\frac{π}{4}, \frac{5π}{4}\right\}$$

B

$$\text{No Solution}$$

C

$$\left\{\frac{7π}{6}, \frac{3π}{2}, \frac{11π}{6}\right\}$$

D

$$\left\{\frac{5π}{6}, \frac{11π}{6}\right\}$$

E

$$\left\{π\right\}$$

Example #3

Solve for 0 ≤ θ < $2π$ $$\frac{-12 - \sqrt{3}}{4}=-3 - \frac{1}{4}\text{tan}\hspace{.1em}θ$$

Please choose the best answer.

A

$$\left\{2\frac{π}{3}, \frac{4π}{3}\right\}$$

B

$$\left\{\frac{3π}{4}, \frac{7π}{6}, \frac{3π}{2}\right\}$$

C

$$\text{No Solution}$$

D

$$\left\{\frac{π}{3}, \frac{5π}{4}, \frac{11π}{12}\right\}$$

E

$$\left\{\frac{π}{3}, \frac{4π}{3}\right\}$$

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