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. 
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: $$\left\{\frac{π}{6}, \frac{π}{3}, \frac{7π}{6}, \frac{4π}{3}\right\}$$
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. $$\text{sin}\hspace{.1em}2θ + \sqrt{2}\text{sin}\hspace{.1em}θ=0$$
Please choose the best answer.
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. $$\sqrt{3}\text{cos}\hspace{.1em}θ + 3\text{sin}\hspace{.1em}2 θ=4\text{sin}\hspace{.1em}2 θ$$
Please choose the best answer.
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. $$-1=3 + 4\text{tan}\left(θ + \frac{π}{6}\right)$$
Please choose the best answer.
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
$$\left\{\frac{7π}{12}, \frac{17π}{12}\right\}$$
Congrats, Your Score is 100%
Better Luck Next Time, Your Score is %
Try again?
Ready for more?
Watch the Step by Step Video Lesson Take the Practice Test
