Solving Trigonometric Equations Part V Lesson

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 for cosine.
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: $$12 \text{cos}^2 β - 9=0$$ At this point, we could solve using factoring or the square root property. Let's use factoring for this example.
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 the double-angle identity for sine.
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: $$\left\{\frac{11π}{12}, \frac{19π}{12}\right\}$$

Solving Trigonometric Equations Using a Sum-to-Product Identity

Example #4: Solve each equation over the interval $[0, π)$.
Usually, the solutions for these problems are given in terms of the general solution. This will be our strategy in the practice test. $$\text{cos}(6x)=-\text{sin}(4x) + \text{cos}(2x)$$ Let's add sin(4x) to both sides and subtract away cos(2x) from both sides. $$\text{cos}(6x) - \text{cos}(2x) + \text{sin}(4x)=0$$ We know from our sum-to-product identities that: $$\text{cos}\hspace{.1em}A - \text{cos}\hspace{.1em}B=-2\text{sin}\left(\frac{A + B}{2}\right)\text{sin}\left(\frac{A - B}{2}\right)$$ $$\text{cos}(6x) - \text{cos}(2x)$$ $$=-2\text{sin}\left(\frac{6x + 2x}{2}\right)\text{sin}\left(\frac{6x - 2x}{2}\right)$$ $$=-2\text{sin}\left(\frac{8x}{2}\right)\text{sin}\left(\frac{4x}{2}\right)$$ $$=-2\text{sin}\left(4x\right)\text{sin}\left(2x\right)$$ Let's replace this in our equation. $$-2\text{sin}\left(4x\right)\text{sin}\left(2x\right) + \text{sin}(4x)=0$$ Let's factor out the sin(4x). $$\text{sin}(4x)(-2\text{sin}\left(2x\right) + 1)=0$$ Use the zero-product property: $$\text{sin}(4x)=0$$ $$\text{or}$$ $$-2\text{sin}(2x) + 1=0$$ Let's begin with the top equation: $$\text{sin}(4x)=0$$ Since we are working from $[0, π)$ let's adjust our interval.
Multiply each part by 4: $$0 ≤ 4x < 4π$$ Where are the sine values 0 in our given interval? $$\text{sin}(0)=0$$ $$\text{sin}(π)=0$$ $$\text{sin}(2π)=0$$ $$\text{sin}(3π)=0$$ This gives us the following: $$4x=0$$ $$4x=π$$ $$4x=2π$$ $$4x=3π$$ Since we have 4x, we can divide each solution by 4. $$x=\frac{0}{4}=0$$ $$x=\frac{π}{4}$$ $$x=\frac{2π}{4}=\frac{π}{2}$$ $$x=\frac{3π}{4}$$ Let's move on to the bottom equation: $$-2\text{sin}(2x) + 1=0$$ Subtract 1 away from each side, then divide both sides by -2: $$\text{sin}(2x)=\frac{1}{2}$$ Since we are working from $[0, π)$ let's adjust our interval.
Multiply each part by 2: $$0 ≤ 2x < 2π$$ Where are the sine values 1/2 in our given interval? $$\text{sin}\left(\frac{π}{6}\right)=\frac{1}{2}$$ $$\text{sin}\left(\frac{5π}{6}\right)=\frac{1}{2}$$ This gives us the following: $$2x=\frac{π}{6}$$ $$2x=\frac{5π}{6}$$ Since we have 2x, we can divide each solution by 2. $$x=\frac{π}{12}$$ $$x=\frac{5π}{12}$$ Let's combine our solutions into one solution set for our interval: $$\left\{0, \frac{π}{12}, \frac{π}{4}, \frac{5π}{12}, \frac{π}{2}, \frac{3π}{4}\right\}$$

---

Skills Check: