Lesson Objectives
• Learn how to solve basic trigonometric equations
• Learn how to solve trigonometric equations with half-angles

## How to Solve Trigonometric Equations with Half-Angles

Over the course of the last few lessons, we learned how to solve trigonometric equations using linear methods, factoring, and squaring. Here, we will take the next step and learn how to work with trigonometric equations that have half-angles.

### Unit Circle

The unit circle will be given here for reference.

### Solving Trigonometric Equations with Half-Angles

In some cases, we will be asked to solve a trigonometric equation with a half-angle. Sometimes, we will need to use our half-angle identities. In other cases, this won't be necessary. Let's look at a few examples.
Example #1: Solve each equation over the interval $[0, 2π)$. $$\text{sin}\hspace{.1em}\frac{β}{2}=\sqrt{2}- \text{sin}\hspace{.1em}\frac{β}{2}$$ Let's first consider our interval written as an inequality: $$0 ≤ β < 2π$$ Since we have β/2, let's divide each part by 2: $$0 ≤ \hspace{.15em}\frac{β}{2}< π$$ Now, let's revisit our equation and find all values of β/2 over the interval $[0, π)$: $$\text{sin}\hspace{.1em}\frac{β}{2}=\sqrt{2}- \text{sin}\hspace{.1em}\frac{β}{2}$$ Add sin β/2 to both sides: $$2\text{sin}\hspace{.1em}\frac{β}{2}=\sqrt{2}$$ Divide both sides by 2: (Note: The angle is β/2) $$\text{sin}\hspace{.1em}\frac{β}{2}=\frac{\sqrt{2}}{2}$$ We want to find all values of β/2 over the interval $[0, π)$ that satisfy our equation. $$\text{sin}\hspace{.1em}\frac{π}{4}=\frac{\sqrt{2}}{2}$$ Sine is also positive in quadrant II.
What angle has a reference angle of $\frac{π}{4}$ or 45° in quadrant II? $$π - \frac{π}{4}=\frac{3π}{4}$$ $$\text{sin}\hspace{.1em}\frac{3π}{4}=\frac{\sqrt{2}}{2}$$ Our last step is to set β/2 equal to each and solve for β. $$\frac{β}{2}=\frac{π}{4}$$ $$\text{or}$$ $$\frac{β}{2}=\frac{3π}{4}$$ Let's start with the top equation. $$\frac{β}{2}=\frac{π}{4}$$ Multiply both sides by 2: $$β=\frac{π}{2}$$ Let's now work on the bottom equation. $$\frac{β}{2}=\frac{3π}{4}$$ Multiply both sides by 2: $$β=\frac{3π}{2}$$ Our solutions for β in our given interval: $$\left\{\frac{π}{2}, \frac{3π}{2}\right\}$$ Let's now look at an example that uses a half-angle identity.
Example #2: Solve each equation over the interval $[0, 2π)$. $$-\text{cos}\hspace{.1em}θ=-2 + 3\text{sin}\hspace{.1em}\frac{θ}{2}$$ Let's replace sin θ/2 using the half-angle identity: $$\text{sin}\hspace{.1em}\frac{A}{2}=\pm \sqrt{\frac{1 - \text{cos A}}{2}}$$ $$-\text{cos}\hspace{.1em}θ=-2 + 3\text{sin}\hspace{.1em}\frac{θ}{2}$$ $$-\text{cos}\hspace{.1em}θ=-2 \pm 3\sqrt{\frac{1 - \text{cos θ}}{2}}$$ Let's add 2 to both sides of the equation: $$2 -\text{cos}\hspace{.1em}θ=\pm 3\sqrt{\frac{1 - \text{cos θ}}{2}}$$ Square both sides: $$(2 -\text{cos}\hspace{.1em}θ)^2=\left(\pm 3\sqrt{\frac{1 - \text{cos}\hspace{.1em}θ}{2}}\right)^2$$ $$4 - 4\text{cos}\hspace{.1em}θ + \text{cos}^2 θ=9 \cdot \frac{1 - \text{cos}\hspace{.1em}θ}{2}$$ Multiply both sides by 2: $$8 - 8\text{cos}\hspace{.1em}θ + 2\text{cos}^2 θ=9(1 - \text{cos}\hspace{.1em}θ)$$ Distribute the 9 on the right-hand side: $$8 - 8\text{cos}\hspace{.1em}θ + 2\text{cos}^2 θ=9 - 9\text{cos}\hspace{.1em}θ$$ Subtract 9 away from each side and add 9 cos θ to both sides: $$-1 + \text{cos}\hspace{.1em}θ + 2\text{cos}^2 θ=0$$ Rearrange into $ax^2 + bx + c=0$: $$2\text{cos}^2 θ + \text{cos}\hspace{.1em}θ - 1=0$$ Factor the left-hand side: $$(2\text{cos}\hspace{.1em}θ - 1)(\text{cos}\hspace{.1em}θ + 1)=0$$ Use the zero-factor property: $$2\text{cos}\hspace{.1em}θ - 1=0$$ $$\text{or}$$ $$\text{cos}\hspace{.1em}θ + 1=0$$ Let's solve the top equation first: $$2\text{cos}\hspace{.1em}θ - 1=0$$ Add 1 to each side, then divide both sides by 2: $$\text{cos}\hspace{.1em}θ=\frac{1}{2}$$ $$θ=\frac{π}{3}, \frac{5π}{3}$$ Let's now solve the bottom equation: $$\text{cos}\hspace{.1em}θ + 1=0$$ $$\text{cos}\hspace{.1em}θ=-1$$ $$θ=π$$ Our solutions for θ in the given interval: $$\left\{\frac{π}{3}, π, \frac{5π}{3}\right\}$$ Note: when we square both sides of an equation, we need to check our solutions in the original equation. To keep this tutorial shorter, we will only show the check when we have an extraneous solution. Let's look at one more example using a half-angle identity.
Example #3: Solve each equation over the interval $[0, 2π)$. $$\sqrt{3}\text{cos}\hspace{.1em}\frac{θ}{2}=1 + \text{cos}\hspace{.1em}θ$$ Let's replace cos θ/2 using the half-angle identity: $$\text{cos}\hspace{.1em}\frac{θ}{2}=\pm \sqrt{\frac{1 + \text{cos}\hspace{.1em}θ}{2}}$$ $$\pm \sqrt{3}\sqrt{\frac{1 + \text{cos}\hspace{.1em}θ}{2}}=1 + \text{cos}\hspace{.1em}θ$$ Square both sides: $$\frac{3(1 + \text{cos}\hspace{.1em}θ)}{2}=1 + 2\text{cos}\hspace{.1em}θ + \text{cos}^2 θ$$ Multiply both sides by 2: $$3(1 + \text{cos}\hspace{.1em}θ)=2 + 4\text{cos}\hspace{.1em}θ + 2 \text{cos}^2 θ$$ Distribute the 3 on the left side: $$3 + 3\text{cos}\hspace{.1em}θ=2 + 4\text{cos}\hspace{.1em}θ + 2 \text{cos}^2 θ$$ Let's move all terms to the right and place in the form: $0=ax^2 + bx + c$: $$0=2\text{cos}^2 θ + \text{cos}\hspace{.1em}θ - 1$$ Flip Sides: $$2\text{cos}^2 θ + \text{cos}\hspace{.1em}θ - 1=0$$ Factor the left-hand side: $$(2\text{cos}\hspace{.1em}θ - 1)(\text{cos}\hspace{.1em}θ + 1)=0$$ Use the zero-product property: $$2\text{cos}\hspace{.1em}θ - 1=0$$ $$\text{or}$$ $$\text{cos}\hspace{.1em}θ + 1=0$$ Let's solve the top equation first: $$2\text{cos}\hspace{.1em}θ - 1=0$$ Add 1 to each side, then divide both sides by 2: $$\text{cos}\hspace{.1em}θ=\frac{1}{2}$$ $$θ=\frac{π}{3}, \frac{5π}{3}$$ Let's now solve the bottom equation: $$\text{cos}\hspace{.1em}θ + 1=0$$ Subtract 1 from each side of the equation: $$\text{cos}\hspace{.1em}θ=-1$$ $$θ=π$$ As we mentioned above and in previous tutorials, when we square both sides of an equation, it is possible to obtain extraneous solutions. Let's check our proposed solutions in the original equation to see if they work: $$\sqrt{3}\text{cos}\hspace{.1em}\frac{θ}{2}=1 + \text{cos}\hspace{.1em}θ$$ Replace θ with each proposed solution: Let's start with $\frac{π}{3}$: $$\sqrt{3}\text{cos}\hspace{.1em}\frac{\large{\frac{π}{3}}}{2}=1 + \text{cos}\hspace{.1em}\frac{π}{3}$$ $$\sqrt{3}\text{cos}\hspace{.1em}\frac{π}{6}=1 + \text{cos}\hspace{.1em}\frac{π}{3}$$ $$\sqrt{3}\cdot \frac{\sqrt{3}}{2}=1 + \frac{1}{2}$$ $$\frac{3}{2}=\frac{2}{2}+ \frac{1}{2}$$ $$\frac{3}{2}=\frac{3}{2}$$ $\frac{π}{3}$ is a valid solution.
Let's now check $\frac{5π}{3}$: $$\sqrt{3}\text{cos}\hspace{.1em}\frac{\large{\frac{5π}{3}}}{2}=1 + \text{cos}\hspace{.1em}\frac{5π}{3}$$ $$\sqrt{3}\text{cos}\hspace{.1em}\frac{5π}{6}=1 + \text{cos}\hspace{.1em}\frac{5π}{3}$$ $$\sqrt{3}\cdot -\frac{\sqrt{3}}{2}=1 + \frac{1}{2}$$ $$-\frac{3}{2}=\frac{2}{2}+ \frac{1}{2}$$ $$-\frac{3}{2}=\frac{3}{2}$$ $\frac{5π}{3}$ is not a valid solution. Lastly, let's check $π$: $$\sqrt{3}\text{cos}\hspace{.1em}\frac{π}{2}=1 + \text{cos}\hspace{.1em}π$$ $$\sqrt{3}\cdot 0=1 + (-1)$$ $$0=0$$ $π$ is a valid solution.

#### Skills Check:

Example #1

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

A
$$\left\{\frac{5π}{6}\right\}$$
B
$$\left\{0\right\}$$
C
$$\left\{π\right\}$$
D
$$\left\{0, \frac{π}{2}\right\}$$
E
$$\left\{\frac{7π}{6}\right\}$$

Example #2

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

A
$$\left\{\frac{π}{3}\right\}$$
B
$$\left\{\frac{2π}{3}, \frac{4π}{3}\right\}$$
C
$$\left\{\frac{π}{3}, \frac{5π}{3}\right\}$$
D
$$\left\{\frac{11π}{6}\right\}$$
E
$$\text{No Solution}$$

Example #3

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

A
$$\left\{\frac{2π}{3}, \frac{4π}{3}\right\}$$
B
$$\left\{\frac{π}{3}, π, \frac{5π}{3}\right\}$$
C
$$\left\{π, \frac{3π}{2}\right\}$$
D
$$\text{No Solution}$$
E
$$\left\{\frac{π}{6}, \frac{11π}{6}\right\}$$