Lesson Objectives
• Learn how to solve basic trigonometric equations
• Learn how to solve trigonometric equations using factoring

## How to Solve Trigonometric Equations Using Factoring

### Unit Circle

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

In the last lesson, we learned the basics of how to solve a trigonometric equation using linear methods. In some cases, we may need to factor in order to solve a trigonometric equation. Let's look at a few examples.
Example #1: Solve each equation over the interval [0°, 360°) and for all solutions. $$3\text{cos}\hspace{.1em}θ + 2\text{cos}^2 θ=-1$$ Add 1 to each side of the equation and rearrange: $$2\text{cos}^2 θ + 3\text{cos}\hspace{.1em}θ + 1=0$$ Factor:
It sometimes helps to think of cos θ as u: $$2u^2 + 3u + 1=0$$ $$(2u + 1)(u + 1)=0$$ $$2\text{cos}^2 θ + 3\text{cos}\hspace{.1em}θ + 1=0$$ $$(2\text{cos}\hspace{.1em}θ + 1)(\text{cos}\hspace{.1em}θ + 1)=0$$ Use the Zero-Product Property: $$2\text{cos}\hspace{.1em}θ + 1=0$$ $$\text{or}$$ $$\text{cos}\hspace{.1em}θ + 1=0$$ Let's start with the top equation: $$2\text{cos}\hspace{.1em}θ + 1=0$$ Subtract 1 away from each side: $$2\text{cos}\hspace{.1em}θ=-1$$ Divide each side by 2: $$\text{cos}\hspace{.1em}θ=-\frac{1}{2}$$ Again, if we reference our unit circle above, we see solutions of 120° and 240°.
If we wanted to use a calculator: $$\text{cos}^{-1}\left(-\frac{1}{2}\right)=120°$$ Note, the reference angle is 180° - 120° = 60°.
What angle in quadrant III has a reference angle of 60°? $$180° + 60°=240°$$ Now, let's work on our bottom equation. $$\text{cos}\hspace{.1em}θ + 1=0$$ Subtract 1 away from each side: $$\text{cos}\hspace{.1em}θ=-1$$ From the unit circle above, we see a single solution of 180°.
Again, we can use our calculator: $$\text{cos}^{-1}(-1)=180°$$ We can put all of our solutions together and obtain: $$\{120°, 180°, 240°\}$$ or using radians: $$\left\{\frac{2π}{3}, π, \frac{4π}{3}\right\}$$ For all solutions, we simply add 360°n to each solution when using degrees or $2π$n when using radians. $$\{120° + 360°n, 180° + 360°n, 240° + 360°n\}$$ or using radians: $$\left\{\frac{2π}{3}+ 2πn, π + 2πn, \frac{4π}{3}+ 2πn\right\}$$ Let's look at another example.
Example #2: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$4=\text{sin}\hspace{.1em}β + 2\text{sin}^2 β + 3$$ Subract 4 away from each side: $$0=\text{sin}\hspace{.1em}β + 2\text{sin}^2 β - 1$$ Switch Sides: $$\text{sin}\hspace{.1em}β + 2\text{sin}^2 β - 1=0$$ Rearrange in the form: $ax^2 + bx + c=0$ $$2\text{sin}^2 β + \text{sin}\hspace{.1em}β - 1=0$$ To factor, we can replace sin β with u. This isn't necessary but it is helpful for some students: $$2u^2 + u - 1=0$$ $$(2u - 1)(u + 1)=0$$ Replace u with sin β: $$(2u - 1)(u + 1)=0$$ $$(2\text{sin}\hspace{.1em}β - 1)(\text{sin}\hspace{.1em}β + 1)=0$$ Solve Using the Zero Product Property: $$2\text{sin}\hspace{.1em}β - 1=0$$ $$\text{or}$$ $$\text{sin}\hspace{.1em}β + 1=0$$ Let's begin with the top equation: $$2\text{sin}\hspace{.1em}β - 1=0$$ Add 1 to each side: $$2\text{sin}\hspace{.1em}β=1$$ Divide each side by 2: $$\text{sin}\hspace{.1em}β=\frac{1}{2}$$ From the unit circle, we see our solutions for β are 30° or 150°. In terms of radians, we get: $$\frac{π}{6}, \frac{5π}{6}$$ Using a calculator, we would find that: $$\text{sin}^{-1}\left(\frac{1}{2}\right)=30°$$ Again, 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. $$180° - 30°=150°$$ Our two solutions for β are 30° and 150°.
Let's look at our other equation: $$\text{sin}\hspace{.1em}β + 1=0$$ Subtract 1 from each side: $$\text{sin}\hspace{.1em}β=-1$$ From the unit circle, we see our solution for β is 270° or in terms of radians: $$\frac{3π}{2}$$ Again, from our calculator: $$\text{sin}^{-1}(-1)=-90°$$ Since we are working from 0 to $2π$, we can add $2π$ or 360° to this to obtain our 270°. We can put all of our solutions together and obtain: $$\{30°, 150°, 270°\}$$ or using radians: $$\left\{\frac{π}{6}, \frac{5π}{6}, \frac{3π}{2}\right\}$$ For all solutions with the sine function, we simply add 360°n to each solution when using degrees or $2π$n when using radians. In this case, notice how going from 30° to 150° to 270° is a jump of 120° each time. This allows us to write our general solution as: $$\{30° + 120°n\}$$ $$\left\{\frac{π}{6}+ \frac{2π}{3}n\right\}$$ Let's look at another example.
Example #3: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$\text{tan}^{2}β + \text{tan}β - 2=0$$ Let's factor the left side: $$(\text{tan}(β) - 1)(\text{tan}(β) + 2)=0$$ Solve Using the Zero Product Property: $$\text{tan}(β) - 1=0$$ $$\text{or}$$ $$\text{tan}(β) + 2=0$$ Let's begin with the top equation. $$\text{tan}(β) - 1=0$$ Add 1 to both sides: $$\text{tan}(β)=1$$ $$\text{tan}^{-1}(1)=\frac{\pi}{4}$$ One solution is $\frac{\pi}{4}$ or 45°. Tangent is positive in quadrants I and III, we can find the angle in quadrant III with a reference angle of $\frac{\pi}{4}$ or 45°. This will be our other solution. $$\frac{4π + \pi}{4}=\frac{5\pi}{4}$$ $$45° + 180°=225°$$ Let's look at our other equation: $$\text{tan}(β) + 2=0$$ Subtract 2 away from each side: $$\text{tan}(β)=-2$$ To keep things simple, let's use the inverse tangent function with positive 2 as the argument. Note: You can also use (-2) as the argument here but you will obtain an angle in quadrant IV rotating clockwise. In this case, you need to add $\pi$ to obtain one solution and $2\pi$ to obtain the other. $$\text{tan}^{-1}(2) ≈ 1.107$$ $$\text{tan}^{-1}(2) ≈ 63.43°$$ The above gives us a reference angle (radians on the top, degrees on the bottom) for our solutions in quadrants II and IV, where tangent is negative.
Quadrant II: $$β=180° - 63.43°=116.57°$$ $$β=\pi - 1.107=2.03$$ Note: the above answers are both approximations. The exact value would be given as: $$β=180° - \text{tan}^{-1}(2)$$ $$β=\pi - \text{tan}^{-1}(2)$$ Quadrant IV: $$β=360° - 63.43°=296.57°$$ $$β=2\pi - 1.107=5.18$$ Note: the above answers are both approximations. The exact value would be given as: $$β=360° - \text{tan}^{-1}(2)$$ $$β=2\pi - \text{tan}^{-1}(2)$$ We can put all of our solutions together and obtain: $$\left\{45°, 116.57°, 225°, 296.57°\right\}$$ or using radians: $$\left\{\frac{π}{4}, 2.03, \frac{5π}{4}, 5.18\right\}$$ For all solutions, the period of tangent is $\pi$ or 180°. $$\left\{\frac{π}{4}+ πn, 2.03 + πn\right\}$$ $$\left\{45° + 180°n, 116.57 + 180°n\right\}$$

### Problems When Dividing Both Sides by a Trigonometric Function

When solving a trigonometric equation, problems can arise when we divide both sides by a trigonometric function. Let's take a look at a simple example.
Example #4: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$\text{cos}β \cdot \text{cot}β=\text{cos}β$$ Let's begin by using factoring to solve the equation.
Subtract cos β away from each side: $$\text{cos}β \cdot \text{cot}β - \text{cos}β=0$$ Factor out cos β: $$\text{cos}β(\text{cot}β - 1)=0$$ Solve Using the Zero Product Property: $$\text{cos}β=0$$ $$\text{or}$$ $$\text{cot}β - 1=0$$ Let's begin with the top equation: $$\text{cos}β=0$$ From the unit circle, we see our solutions for β are 90° or 270°. In terms of radians, we get: $$\frac{\pi}{2}, \frac{3\pi}{2}$$ Let's look at our other equation: $$\text{cot}β - 1=0$$ Add 1 to each side: $$\text{cot}β=1$$ Use the reciprocal identity: $$\text{tan}β=\frac{1}{\text{cot}β}$$ $$\text{tan}β=1$$ Use the inverse tangent function. $$\text{tan}^{-1}(1)=\frac{\pi}{4}$$ We know the period of tangent is $\pi$ or 180°. This means we would also have a solution in quadrant III. $$\pi + \frac{\pi}{4}=\frac{4\pi + \pi}{4}=\frac{5\pi}{4}$$ Let's put all of our solutions together and obtain: $$\left\{\frac{\pi}{4}, \frac{\pi}{2}, \frac{5\pi}{4}, \frac{3\pi}{2}\right\}$$ or using degrees: $$\left\{45°, 90°, 225°, 270°\right\}$$ For all solutions: $$\left\{\frac{\pi}{4}+ πn, \frac{\pi}{2}+ πn\right\}$$ or using degrees: $$\left\{45° + 180°n, 90° + 180°n\right\}$$ In this case, it may be helpful to look at the solutions using a basic sketch. Let's try to solve the problem using a different approach. Let's suppose we decided to divide both sides of the equation by cos β. $$\text{cos}β \cdot \text{cot}β=\text{cos}β$$ $$\frac{\text{cos}β \cdot \text{cot}β}{\text{cos}β}=\frac{\text{cos}β}{\text{cos}β}$$ $$\text{cot}β=1$$ Here, we would only get that β is 45° or 225° ($\frac{\pi}{4}$ or $\frac{5\pi}{4}$). What happened to the solutions of 90° and 270° ($\frac{\pi}{2}$ and $\frac{3\pi}{2}$) we found earlier? We are missing the solutions from the original equation that make the divisor zero. $$\text{cos}\frac{\pi}{2}=0$$ $$\text{cos}\frac{3\pi}{2}=0$$ For this reason, it is recommended that we don't divide both sides by a trigonometric function such as cos β.

#### Skills Check:

Example #1

Solve each equation for 0 ≤ θ < 2π $$\text{cos}\hspace{.1em}θ + \sqrt{2}\text{cos}\hspace{.1em}θ \hspace{.1em}\text{sin}\hspace{.1em}θ=0$$

A
$$\left\{0, \frac{π}{4}\right\}$$
B
$$\left\{0, π, \frac{5π}{6}\right\}$$
C
$$\left\{\frac{π}{2}, \frac{5π}{4}, \frac{3π}{2}, \frac{7π}{4}\right\}$$
D
$$\left\{\frac{2π}{3}\right\}$$
E
$$\text{No Solution}$$

Example #2

Solve each equation for 0 ≤ β < 2π $$2\sqrt{3}\text{sin}\hspace{.1em}β \hspace{.1em}\text{cos}\hspace{.1em}β=3\text{sin}\hspace{.1em}β$$

A
$$\left\{0, \frac{π}{6}, π, \frac{11π}{6}\right\}$$
B
$$\left\{\frac{π}{2}, \frac{3π}{2}\right\}$$
C
$$\left\{π, \frac{11π}{6}\right\}$$
D
$$\left\{0, \frac{π}{6}\right\}$$
E
$$\text{No Solution}$$

Example #3

Solve each equation for 0 ≤ θ < 2π $$2\sqrt{3}\text{tan}\hspace{.1em}θ \text{cos}\hspace{.1em}θ + 3 \text{tan}\hspace{.1em}θ=0$$

A
$$\text{No Solution}$$
B
$$\left\{0, \frac{π}{3}, \frac{2π}{3}\right\}$$
C
$$\left\{0, \frac{5π}{6}, π\right\}$$
D
$$\left\{\frac{11π}{6}\right\}$$
E
$$\left\{0, \frac{5π}{6}, π, \frac{7π}{6}\right\}$$         