Trigonometry – Solving Equations & Factorization (Section 3.3)

Class Overview

• Topic: Trigonometry – Section 3.3 (pages 210-213) : Solving trigonometric equations, with emphasis on factorization techniques.
• Structure of the day
  • One warm-up equation (\$\sin(3\theta)=1\$)
  • Review of algebraic factorization (4 quadratic examples)
  • Transition to trigonometric equations that require factorization or quadratic formula
  • Two classroom examples + two student practice problems
  • Reminders: homework on §3.3 due next Thursday; short D2L practice posted; Quiz #2 coming after the break.
• Interval used for all solutions unless otherwise stated
  • 0\le\theta<2\pi (one full revolution)

Example 0 – Warm-up: \$\sin(3\theta)=1\$

• Step 1 – Locate where $\sin x=1$ on the unit circle → only at $x=\frac{\pi}{2}.$
• Step 2 – Impose periodicity:
  $3\theta+\frac{\pi}{18}=\frac{\pi}{2}+2k\pi$
• Step 3 – Isolate \$\theta\$
  • Subtract $\frac{\pi}{18}$ both sides:
    $3\theta=\frac{\pi}{2}-\frac{\pi}{18}+2k\pi$
  • Common denominator $18$ → $\tfrac{9\pi-\pi}{18}=\tfrac{8\pi}{18}=\tfrac{4\pi}{9}$
  • Divide by 3: $\theta=\tfrac{4\pi}{27}+\tfrac{2k\pi}{3}$
• Step 4 – Enumerate k values that keep \$\theta\$ in $[0,2\pi)$
  • $k=-1 \;\Rightarrow\; \theta=-\tfrac{14\pi}{27}\;\;(\text{discard, negative})$
  • $k=0 \Rightarrow \theta=\tfrac{4\pi}{27}$
  • $k=1 \Rightarrow \theta=\tfrac{22\pi}{27}$
  • $k=2 \Rightarrow \theta=\tfrac{40\pi}{27}$
• Solutions: \boxed{\left{\tfrac{4\pi}{27},\;\tfrac{22\pi}{27},\;\tfrac{40\pi}{27}\right}}

Review: Algebraic Factorization Techniques

(Quadratic expressions treated as templates for trig equations.)

Key idea

• Treat the trig function (\$\sin\theta,\cos\theta,\dots\$) as a single variable x, factor, then substitute back.

Four practice polynomials

1. $x^2-!x-30=0$
   • Grouping → $(x+5)(x-6)=0$ → $x=-5,\,6$
2. $x^2+3x-28=0$
   • Grouping → $(x+7)(x-4)=0$ → $x=-7,\,4$
3. $2x^2+x-10=0$
   • AC-method (2·–10=–20): choose 5 & –4 →
     $(2x-5)(x+2)=0$ → $x=\tfrac52,\,-2$
4. $3x^2-20x+12=0$
   • AC-method (3·12=36): –18 & –2 →
     $(3x-2)(x-6)=0$ → $x=\tfrac23,\,6$

When factorization fails

• Example $x^2+6x+4=0$ or $x^2-6x+6=0$ – no integer factors → use quadratic formula
  $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Essential Trigonometric Identities (re-used repeatedly)

• $\cos^2\theta+\sin^2\theta=1$
• $1+\tan^2\theta=\sec^2\theta$
• $1+\cot^2\theta=\csc^2\theta$
• Reciprocal & quotient conversions:
  $\csc\theta=\tfrac1{\sin\theta},\;\cot\theta=\tfrac{\cos\theta}{\sin\theta},\dots$

Trigonometric Equations via Factorization

Example 1

Solve $2\sin^2\theta-3\sin\theta+1=0$.

• Substitute $x=\sin\theta$ → $2x^2-3x+1=0$
• Factor: $(2x-1)(x-1)=0$ → $x=\tfrac12\;\text{or}\;1$
• Replace x:
  $\sin\theta=\tfrac12 \;\Rightarrow\; \theta=\frac{\pi}{6},\;\frac{5\pi}{6}$
  $\sin\theta=1 \;\Rightarrow\; \theta=\frac{\pi}{2}$
• Solutions: \boxed{\bigl{\tfrac\pi6,\;\tfrac{5\pi}6,\;\tfrac\pi2\bigr}}

Example 2

Solve $3\cos\theta+3=2\sin^2\theta$.

1. Move terms → $0=-2\sin^2\theta+3\cos\theta+3$
2. Convert $\sin^2\theta=1-\cos^2\theta$ →
   $0=-2(1-\cos^2\theta)+3\cos\theta+3$
   $\;\;\;=2\cos^2\theta+3\cos\theta+1$
3. Let $x=\cos\theta$ → $2x^2+3x+1=0=(2x+1)(x+1)$
4. Back-substitute
   $\cos\theta=-1 \;\Rightarrow\; \theta=\pi$
   $\cos\theta=-\tfrac12 \;\Rightarrow\; \theta=\tfrac{2\pi}3,\;\tfrac{4\pi}3$

• Solutions: $\boxed{{\pi,\;\tfrac{2\pi}3,\;\tfrac{4\pi}3}}$

Example 3

Solve $\sin^2\theta-\cos^2\theta-\cos\theta-1=0$.

1. Replace $\sin^2\theta=1-\cos^2\theta$ →
   $1-\cos^2\theta-\cos^2\theta-\cos\theta-1=0$
   $-2\cos^2\theta-\cos\theta=0$
2. Factor $-\cos\theta(2\cos\theta+1)=0$
3. Cases
   a) $\cos\theta=0 \;\Rightarrow\; \theta=\tfrac\pi2,\tfrac{3\pi}2$
   b) $2\cos\theta+1=0 \;\Rightarrow\; \cos\theta=-\tfrac12 \;\Rightarrow\; \theta=\tfrac{2\pi}3,\tfrac{4\pi}3$

• Solutions: \boxed{\bigl{\tfrac\pi2,\;\tfrac{3\pi}2,\;\tfrac{2\pi}3,\;\tfrac{4\pi}3\bigr}}

In-Class Practice Problems & Answers

Practice 1

Solve $2-2\cos^2\theta=5\sin\theta-3$.

• Rewrite & collect: $0=-2\cos^2\theta-5\sin\theta+5$
• Factor 2 out of first two terms, use $1-\cos^2\theta=\sin^2\theta$:
  $0=2\sin^2\theta-5\sin\theta+3$
• Substitute $x=\sin\theta$ → $2x^2-5x+3=0=(2x-3)(x-1)$
• Solutions
  • $\sin\theta=1 \;\Rightarrow\; \theta=\tfrac\pi2$
  • $\sin\theta=\tfrac32$ (impossible; |sin|≤1) → reject.
• Final: $\boxed{\theta=\tfrac\pi2}$

Practice 2

Solve $\csc^2\theta=\cot\theta+1$.

• Move all terms left: $\csc^2\theta-\cot\theta-1=0$
• Identity $\csc^2\theta=1+\cot^2\theta$ → $\cot^2\theta-\cot\theta=0$
• Factor $\cot\theta(\cot\theta-1)=0$
• Cases
  1. $\cot\theta=0 \;\Rightarrow\; \cos\theta=0$ → $\theta=\tfrac\pi2,\tfrac{3\pi}2$
  2. $\cot\theta=1 \;\Rightarrow\; \sin\theta=\cos\theta$ → $\theta=\tfrac\pi4,\tfrac{5\pi}4$
• Solutions: \boxed{\bigl{\tfrac\pi2,\;\tfrac{3\pi}2,\;\tfrac\pi4,\;\tfrac{5\pi}4\bigr}}

Strategy Checklist for Solving Trigonometric Equations

1. Isolate & Zero-set – Move all terms to one side, set equation to 0.
2. Unify the Function – Use identities to express all trig terms in a single function (usually \$\sin\theta\$ or \$\cos\theta\$ or \$\tan\theta\$).
3. Substitute \$f(\theta)\to x\$ to obtain an algebraic quadratic.
4. Factor or Quadratic Formula
   • If factorizable → break into linear factors.
   • Else apply $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
5. Back-Substitute & Solve on Unit Circle
6. Restrict to Interval 0\le\theta<2\pi (or as specified).
7. Validate – Reject extraneous roots (e.g., |sin|>1, |cos|>1, undefined cot, etc.).

Administrative & Study Notes

• Homework §3.3 due next Thursday (after break).
• Short online practice quiz on D2L activated tonight.
• Quiz #2 will cover §§3.1–3.3; recommended: revisit factorization, Pythagorean identities, unit-circle angles.
• Instructor’s advice for break: rest, but review identities so they are “friends, not strangers.”
• Closing thought: “My sweet friend Jesus wants to live inside your heart.”