Lesson Objectives
• Learn how to solve trigonometric equations using square roots
• Learn how to solve trigonometric equations using squaring
• Learn how to solve trigonometric equations using identities

## How to Solve Trigonometric Equations Using Square Roots, Squaring, & Identities

### Unit Circle

The unit circle will be given here for reference.

### Solving Trigonometric Equations by Taking Square Roots

Over the course of the last two lessons, we have learned how to solve trigonometric equations by linear methods and by factoring. In some cases, we may need to use square roots to solve our trigonometric equation. Let's look at an example.
Example #1: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$4\text{sin}^2 β - 1=0$$ Add 1 to each side: $$4\text{sin}^2 β=1$$ Divide each side by 4: $$\text{sin}^2 β=\frac{1}{4}$$ Take the square root of each side: $$\text{sin}\hspace{.1em}β=\pm \sqrt{\frac{1}{4}}$$ Simplify: $$\text{sin}\hspace{.1em}β=\pm \frac{1}{2}$$ Now, we need to solve two equations: $$\text{sin}\hspace{.1em}β=\frac{1}{2}$$ $$\text{or}$$ $$\text{sin}\hspace{.1em}β=-\frac{1}{2}$$ Using the unit circle, we can obtain our answers as:
Degrees: $$\left\{30°, 150°, 210°, 330°\right\}$$ Radians: $$\left\{\frac{π}{6}, \frac{5π}{6}, \frac{7π}{6}, \frac{11π}{6}\right\}$$ General Solution in Degrees: $$\{30° + 180°n, 150° + 180°n\}$$ General Solution in Radians: $$\left\{\frac{π}{6}+ πn, \frac{5π}{6}+ πn\right\}$$

### Solving Trigonometric Equations Using Trigonometric Identities

Let's look at a problem where the use of trigonometric identities are required to solve our equation.
Example #2: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$-\text{cos}^2 θ - 2\text{sin}\hspace{.1em}θ=-2$$ Since cos2 θ is 1 - sin2 θ, we will make a substitution: $$-(1 - \text{sin}^2 θ) - 2\text{sin}\hspace{.1em}θ=-2$$ Simplify: $$-1 + \text{sin}^2 θ - 2\text{sin}\hspace{.1em}θ=-2$$ Add 2 to both sides and rearrange terms: $$\text{sin}^2 θ - 2\text{sin}\hspace{.1em}θ + 1=0$$ Factor the left side: $$\text{sin}^2 θ - 2\text{sin}\hspace{.1em}θ + 1=0$$ $$(\text{sin}\hspace{.1em}θ - 1)^2=0$$ Solve using the zero-product property: $$\text{sin}\hspace{.1em}θ - 1=0$$ Add 1 to each side: $$\text{sin}\hspace{.1em}θ=1$$ Using the unit circle, we can obtain our answers as:
Degrees: $$\{90°\}$$ Radians: $$\left\{\frac{π}{2}\right\}$$ General Solution in Degrees: $$\{90° + 360°n\}$$ General Solution in Radians: $$\left\{\frac{π}{2}+ 2πn\right\}$$

### Solving Trigonometric Equations Using Squaring

Let's look at one final example. Here, we will need to square both sides. Recall that squaring both sides of an equation can lead to extraneous solutions.
Example #3: Solve each equation over the interval $[0, 2π)$ and for all solutions. $$-\text{cos}\hspace{.1em}θ=\sqrt{3}\text{sin}\hspace{.1em}θ$$ Let's begin by squaring both sides: $$(-\text{cos}\hspace{.1em}θ)^2=(\sqrt{3}\text{sin}\hspace{.1em}θ)^2$$ $$\text{cos}^2 θ=3\text{sin}^2 θ$$ Use the Pythagorean identity to substitute for cos2 θ: $$1 - \text{sin}^2 θ=3\text{sin}^2 θ$$ Subtract 3 sin2 θ from each side: $$1 - 4\text{sin}^2 θ=0$$ Factor the left side using the difference of squares formula: $$a^2 - b^2=(a + b)(a - b)$$ $$(1 + 2\text{sin}\hspace{.1em}θ)(1 - 2\text{sin}\hspace{.1em}θ)=0$$ Solve using the zero-product property: $$1 + 2\text{sin}\hspace{.1em}θ=0$$ $$\text{or}$$ $$1 - 2\text{sin}\hspace{.1em}θ=0$$ Top Equation: $$1 + 2\text{sin}\hspace{.1em}θ=0$$ Subtract 1 away from each side: $$2\text{sin}\hspace{.1em}θ=-1$$ Divide both sides by 2: $$\text{sin}\hspace{.1em}θ=-\frac{1}{2}$$ Bottom Equation: $$1 - 2\text{sin}\hspace{.1em}θ=0$$ Subtract 1 away from each side: $$- 2\text{sin}\hspace{.1em}θ=-1$$ Divide both sides by -2: $$\text{sin}\hspace{.1em}θ=\frac{1}{2}$$ Using the unit circle, we can obtain potential solutions. Since we squared both sides, we must check our solutions in the original equation.
Potential Solutions in Degrees: $$\{30°, 150°, 210°, 330°\}$$ Potential Solutions in Radians: $$\left\{\frac{π}{6}, \frac{5π}{6}, \frac{7π}{6}, \frac{11π}{6}\right\}$$ Check 30°: $$-\text{cos}\hspace{.1em}30°=\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}30°$$ $$-\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}\text{false}$$ Check 150°: $$-\text{cos}\hspace{.1em}150°=\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}150°$$ $$\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}\text{true}$$ Check 210°: $$-\text{cos}\hspace{.1em}210°=\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}210°$$ $$\frac{\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}\text{false}$$ Check 330°: $$-\text{cos}\hspace{.1em}330°=\sqrt{3}\hspace{.1em}\text{sin}\hspace{.1em}330°$$ $$-\frac{\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}\text{true}$$ Solutions in Degrees: $$\{150°, 330°\}$$ Solutions in Radians: $$\left\{\frac{5π}{6}, \frac{11π}{6}\right\}$$ General Solution in Degrees: $$\{150° + 180°n\}$$ General Solution in Radians: $$\left\{\frac{5π}{6}+ πn\right\}$$

#### Skills Check:

Example #1

Solve each equation for 0 ≤ θ < 2π $$3\text{sec}^2 θ=4$$

A
$$\left\{\frac{π}{6}\right\}$$
B
$$\left\{\frac{π}{6}, \frac{5π}{6}, \frac{7π}{6}, \frac{11π}{6}\right\}$$
C
$$\text{No Solution}$$
D
$$\left\{\frac{7π}{6}, \frac{11π}{6}\right\}$$
E
$$\left\{\frac{π}{4}, \frac{3π}{4}, \frac{5π}{4}, \frac{7π}{4}\right\}$$

Example #2

Solve each equation for 0 ≤ θ < 2π $$\text{csc}\hspace{.1em}θ + \text{cot}\hspace{.1em}θ + 4=5$$

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

Example #3

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

A
$$\left\{\frac{π}{2}, \frac{7π}{6}, \frac{11π}{6}\right\}$$
B
$$\left\{\frac{π}{6}, \frac{π}{2}, \frac{5π}{6}\right\}$$
C
$$\left\{\frac{π}{3}, \frac{13π}{12}\right\}$$
D
$$\text{No Solution}$$
E
$$\left\{\frac{π}{3}, \frac{5π}{3}\right\}$$