Section 7.5

Solving Trigonometric Equations

This is the capstone of Chapters 5–7 and the most heavily tested topic on Exam 3. Every concept — unit circle values, identities, double-angle formulas, inverse trig — comes together here. If you can solve trig equations confidently, you've got the whole exam covered.

⚡ Quick Summary

Solving a trig equation is like a detective hunt — you're looking for which angles make the equation true. The simplest case: $\sin(x) = \frac{1}{2}$ asks "where on the unit circle does sine equal $\frac{1}{2}$ ?" But here's the twist — trig functions repeat forever, so there are infinitely many answers. Usually the exam asks for solutions in one full trip around the circle, $[0, 2\pi)$ , or the general solution where you add $+ 2n\pi$ to capture every repetition.

General Solution (Sine/Cosine)

x = x_0 + 2n\pi, \quad n \in \mathbb{Z}

Sine and cosine repeat every 2π.

General Solution (Tangent)

x = x_0 + n\pi, \quad n \in \mathbb{Z}

Tangent repeats every π.

📋 Before You Start

Tap any item if you need a refresher:

Plain-English version

For harder equations, the universal strategy is: get everything down to one trig function, then solve. You might need to use identities (like replacing $\cos(2x)$ with $1 - 2\sin^2(x)$ ), factor a quadratic (treat $\sin(x)$ like a variable), or apply formulas from earlier sections. Think of all those identities you learned as your toolbox — this is where you actually use them.

Two rules will save you from the most common mistakes. First: never divide by a trig expression — factor instead. If you divide both sides by $\sin(x)$ , you lose every solution where $\sin(x) = 0$ . Second: when you take a square root, **don't forget the $\pm$ **. If $\sin^2(x) = \frac{1}{4}$ , then $\sin(x) = \frac{1}{2}$ OR $\sin(x) = -\frac{1}{2}$ , giving four solutions instead of two.

This section is the grand finale — it pulls together unit circle values, identities, double-angle formulas, and inverse trig. Use the F.I.R.S.T. checklist: Factor (don't divide), Isolate the trig function, Reference angle from the unit circle, Sign determines which quadrants, Test that your answers are in the requested interval.

Key Formulas

General Solution (Sine/Cosine)

x = x_0 + 2n\pi, \quad n \in \mathbb{Z}

Sine and cosine repeat every 2π.

General Solution (Tangent)

x = x_0 + n\pi, \quad n \in \mathbb{Z}

Tangent repeats every π.

Zero Product Property

AB = 0 \implies A = 0 \text{ or } B = 0

Factor and set each factor to zero — never divide by a trig expression.

Quadratic Substitution

a\sin^2(x) + b\sin(x) + c = 0 \implies \text{let } u = \sin(x)

Treat it like au² + bu + c = 0, solve for u, then find x.

Key Takeaways

✓Strategy: get everything down to one trig function, then solve — use identities, factoring, or substitution
✓Never divide by a trig expression — factor instead, or you'll lose solutions where that expression equals zero
✓Don't forget $\pm$ when taking a square root — $\sin^2 x = \frac{1}{4}$ gives $\sin x = \pm\frac{1}{2}$
✓General solutions: add $+ 2n\pi$ for sin/cos equations, $+ n\pi$ for tangent equations ( $n \in \mathbb{Z}$ )

⚠️ Common Mistakes

✗Dividing both sides by $\sin(x)$ or $\cos(x)$ instead of factoring — this loses solutions where that function equals zero
✗Forgetting the $\pm$ when taking a square root: $\sin^2(x) = \frac{1}{4}$ means $\sin(x) = \frac{1}{2}$ OR $\sin(x) = -\frac{1}{2}$ (four solutions, not two)
✗Only checking one or two quadrants — after finding the reference angle, check ALL quadrants where the trig function has the correct sign
✗Not verifying that your answers are in the requested interval $[0, 2\pi)$ — always do a final check
✗Forgetting to use identities to reduce to one trig function — if you see $\cos(2x)$ , replace it with a double-angle formula first

Builds on: 7.1 Builds on: 7.2 Builds on: 7.3 Builds on: 6.3

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7.3 Double-Angle & Half-Angle Formulas