Lesson Objectives
• Learn how to solve basic trigonometric equations
• Learn how to solve trigonometric equations with steps inside
• Learn how to solve trigonometric equations with multiple angles

## How to Solve Trigonometric Equations with Multiple Angles

Here, we will take the next step and learn how to work with trigonometric equations that have multiple angles and/or steps inside.

### Unit Circle

The unit circle will be given here for reference.

### Solving Trigonometric Equations with Multiple Angles

In some cases, we will need to solve trigonometric equations with multiple angles. Let's begin by solving an equation using a double-angle identity.
Example #1: Solve each equation over the interval $[0, 2π)$. $$\text{cos}\hspace{.1em}2β=-10 \text{cos}^2 β + 8$$ Since we have 2β, let's use our double-angle identity for cosine: $$\text{cos}2β=2 \text{cos}^2 β - 1$$ Let's perform a substitution in the original equation. $$2 \text{cos}^2 β - 1=-10 \text{cos}^2 β + 8$$ Add $10 \text{cos}^2 β$ to both sides: $$12 \text{cos}^2 β - 1=+ 8$$ Subtract 8 from both sides: $$12 \text{cos}^2 β - 9=0$$ Factor out a 3: $$3(4 \text{cos}^2 β - 3)=0$$ In this case, it may not be obvious that we can factor this using the difference of squares formula. $$3((2 \text{cos}\hspace{.1em}β)^2 - (\sqrt{3})^2)=0$$ $$3(2 \text{cos}\hspace{.1em}β + \sqrt{3})(2 \text{cos}\hspace{.1em}β - \sqrt{3})=0$$ Use the zero-product property: $$(2 \text{cos}\hspace{.1em}β + \sqrt{3})=0$$ $$\text{or}$$ $$(2 \text{cos}\hspace{.1em}β - \sqrt{3})=0$$ Let's solve the top equation first. $$(2 \text{cos}\hspace{.1em}β + \sqrt{3})=0$$ Subtract $\sqrt{3}$ away from each side, then divide each side by 2: $$\text{cos}\hspace{.1em}β=-\frac{\sqrt{3}}{2}$$ From the unit circle our solutions are: $$β=\frac{5π}{6}, \frac{7π}{6}$$ Let's now solve the bottom equation. $$(2 \text{cos}\hspace{.1em}β - \sqrt{3})=0$$ Add $\sqrt{3}$ to each side, then divide each side by 2: $$\text{cos}\hspace{.1em}β=\frac{\sqrt{3}}{2}$$ From the unit circle our solutions are: $$β=\frac{π}{6}, \frac{11π}{6}$$ Let's combine our solutions into one solution set for our interval: $$\left\{\frac{π}{6}, \frac{5π}{6}, \frac{7π}{6}, \frac{11π}{6}\right\}$$ Let's now look at an example using a multiple-angle identity.
Example #2: Solve each equation over the interval $[0, 2π)$. $$4\text{sin}\hspace{.1em}α \hspace{.1em}\text{cos}\hspace{.1em}α=\sqrt{3}$$ Since we see both sine and cosine in our equation, let's think about a way to rewrite this equation: $$2(2\text{sin}\hspace{.1em}α \hspace{.1em}\text{cos}\hspace{.1em}α)=\sqrt{3}$$ We know from our double-angle identities that: $$2\hspace{.1em}\text{sin}\hspace{.1em}α \hspace{.1em}\text{cos}\hspace{.1em}α=\text{sin}\hspace{.1em}2 α$$ Let's perform a substitution in the original equation. $$2\text{sin}\hspace{.1em}2 α=\sqrt{3}$$ Divide both sides by 2: $$\text{sin}\hspace{.1em}2 α=\frac{\sqrt{3}}{2}$$ From here, consider the fact that we are working with $2α$: $$0 ≤ α < 2π$$ Multiply all parts by 2: $$0 ≤ 2α < 4π$$ When we think about $\frac{\sqrt{3}}{2}$:
In Quadrant I: $$\text{sin}\frac{π}{3}=\frac{\sqrt{3}}{2}$$ In Quadrant II: $$\text{sin}\hspace{.1em}\frac{2π}{3}=\frac{\sqrt{3}}{2}$$ Since we are going out to $4π$ we need to add $2π$ to each answer: $$\frac{π}{3}+ \frac{6π}{3}=\frac{7π}{3}$$ $$\frac{2π}{3}+ \frac{6π}{3}=\frac{8π}{3}$$ $$2 α=\frac{π}{3}, \frac{2π}{3}, \frac{7π}{3}, \frac{8π}{3}$$ Since we want to solve for $α$, we can divide all solutions by 2: $$α=\frac{π}{6}, \frac{π}{3}, \frac{7π}{6}, \frac{4π}{3}$$ Let's combine our solutions into one solution set for our interval: $$\left\{\frac{π}{6}, \frac{π}{3}, \frac{7π}{6}, \frac{4π}{3}\right\}$$

### Solving Trigonometric Equations with Steps Inside

Example #3: Solve each equation over the interval $[0, 2π)$. $$-\frac{9}{2}=-4 + \text{cos}\left(β + \frac{7π}{4}\right)$$ First, let's isolate our trigonometric expression. Add 4 to each side: $$-\frac{9}{2}+ 4=\text{cos}\left(β + \frac{7π}{4}\right)$$ $$-\frac{9}{2}+ \frac{8}{2}=\text{cos}\left(β + \frac{7π}{4}\right)$$ $$-\frac{1}{2}=\text{cos}\left(β + \frac{7π}{4}\right)$$ Switch Sides: $$\text{cos}\left(β + \frac{7π}{4}\right)=-\frac{1}{2}$$ From here, consider the fact that we are working with $β + \frac{7π}{4}$: $$0 ≤ β < 2π$$ Add $\frac{7π}{4}$ to all parts: $$\frac{7π}{4}≤ β + \frac{7π}{4}< \frac{15π}{4}$$ Let's again think about where the cosine values are -1/2 in our given interval: $$\text{cos}\hspace{.1em}\frac{8π}{3}=-\frac{1}{2}$$ $$\text{cos}\hspace{.1em}\frac{10π}{3}=-\frac{1}{2}$$ Since we want to solve for β, we can subtract $\frac{7π}{4}$ from each result. $$β + \frac{7π}{4}=\frac{8π}{3}$$ $$\text{or}$$ $$β + \frac{7π}{4}=\frac{10π}{3}$$ Top Equation: $$β=\frac{8π}{3}- \frac{7π}{4}$$ $$β=\frac{32π}{12}- \frac{21π}{12}$$ $$β=\frac{11π}{12}$$ Bottom Equation: $$β=\frac{10π}{3}- \frac{7π}{4}$$ $$β=\frac{40π}{12}- \frac{21π}{12}$$ $$β=\frac{19π}{12}$$ Let's combine our solutions into one solution set for our interval: $$\left\{\frac{11π}{12}, \frac{19π}{12}\right\}$$

#### Skills Check:

Example #1

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

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

Example #2

Solve each for 0 ≤ θ < 2π. $$\sqrt{3}\text{cos}\hspace{.1em}θ + 3\text{sin}\hspace{.1em}2 θ=4\text{sin}\hspace{.1em}2 θ$$

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

Example #3

Solve each for 0 ≤ θ < 2π. $$-1=3 + 4\text{tan}\left(θ + \frac{π}{6}\right)$$

A
$$\left\{\frac{7π}{12}, \frac{19π}{12}\right\}$$
B
$$\text{No Solution}$$
C
$$\left\{\frac{5π}{12}, \frac{7π}{12}\right\}$$
D
$$\left\{\frac{5π}{12}, \frac{19π}{12}\right\}$$
E
$$\left\{\frac{7π}{12}, \frac{17π}{12}\right\}$$

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