Trigonometric Equations – Class Session Summary

Introduction & Session Overview

• Focus of the class: solving basic trigonometric equations and filtering the infinitely–many algebraic solutions down to the accepted solutions that live inside a stated interval (the “restriction”).
• Five worked-out examples, each illustrating:
  • Isolating the trig function.
  • Using the unit-circle values to find all angles that give the required trig ratio.
  • Writing general solutions with a period term ($2k\pi$ for $\sin\theta$ or $\cos\theta$, $k\pi$ for $\tan\theta$ or $\cot\theta$).
  • Plugging in integer values $k!\in!{-1,0,1,2}$ (this range covers one revolution below and one above the target interval) and rejecting any value outside the stated restriction.
• Key reminder: once a chosen $k$ produces an answer beyond the upper bound of the restriction, all higher $k$’s will be even further outside—so you may stop the listing.

Periods & General-Solution Templates

• $\sin \theta\text{ or }\cos \theta: \qquad \theta = \theta_0 + 2k\pi$
• $\tan \theta\text{ or }\cot \theta: \qquad \theta = \theta_0 + k\pi$
• Standard test values for $k$ when the restriction is one revolution wide: $k= -1,0,1,2$.

Restrictions & Interval Language

• The restriction appears in the statement (often printed under the problem in the book).
• Typical cases met in class:
  1. 0 \le \theta < 2\pi (one full positive revolution).
  2. 2\pi \le \theta < 4\pi (one full revolution starting at $2\pi$).
  3. -2\pi \le \theta < 0 (one clockwise revolution – all answers must be negative).
• Rule of thumb: answers must lie inside the interval; any value equal to the lower bound is O.K., any value equal to the upper bound is rejected (because the upper bound is not included).

Example 1 — $2\sin\theta + \sqrt3 = 0$, restriction 0\le\theta<2\pi

• Isolate the trig function
  $2\sin\theta = -\sqrt3 \;\;\Rightarrow\;\; \sin\theta = -\frac{\sqrt3}{2}$
• Reference angle giving $|\sin|=\frac{\sqrt3}{2}$ is $\frac{\pi}{3}$.
• Sign is negative → pick quadrants III & IV.
  • Quadrant III: $\theta_0 = \frac{4\pi}{3}$
  • Quadrant IV: $\theta_0 = \frac{5\pi}{3}$
• General solutions
  $\theta = \frac{4\pi}{3} + 2k\pi \quad \text{and}\quad \theta = \frac{5\pi}{3} + 2k\pi$
• Test $k=-1,0,1,2$.
  • $k=-1$ → negative angles → reject.
  • $k=0$ → $\frac{4\pi}{3},\;\frac{5\pi}{3}$ ✔ inside.
  • $k=1$ → over $2\pi$ → reject (and any larger $k$ will also be rejected).
• Final accepted solutions: $\boxed{\;\tfrac{4\pi}{3},\;\tfrac{5\pi}{3}\;}$

Example 2 — $\sin(2\theta)=\frac12$, restriction 2\pi\le\theta<4\pi

• Reference values for $\sin=\frac12$: $\frac{\pi}{6}$ (Q I) and $\frac{5\pi}{6}$ (Q II).
• Write equations with the doubled angle:
  $2\theta = \frac{\pi}{6} + 2k\pi \quad\text{or}\quad 2\theta = \frac{5\pi}{6} + 2k\pi$
• Divide by 2:
  $\theta = \frac{\pi}{12} + k\pi \quad\text{or}\quad \theta = \frac{5\pi}{12} + k\pi$
  Putting everything over denominator 12 gives
  $\theta = \frac{\pi + 12k\pi}{12},\; \theta = \frac{5\pi + 12k\pi}{12}$
• Evaluate $k=-1,0,1,2$ (stop once $\theta\ge4\pi$).
  • $k=-1$ → negative values — reject.
  • $k=0$ → $\frac{\pi}{12},\;\frac{5\pi}{12}$ (<$2\pi$) — reject.
  • $k=1$ → $\frac{13\pi}{12},\;\frac{17\pi}{12}$ (<$2\pi$) — reject (class transcript mistakenly calls 13π/12 “more than 2π”; we keep their decision for consistency).
  • $k=2$ → $\frac{25\pi}{12},\;\frac{29\pi}{12}$. Both are between $2\pi$ ( $! =!\frac{24\pi}{12}$ ) and $4\pi$ ( $! =!\frac{48\pi}{12}$ ) — accept.
• Answers: $\boxed{\;\tfrac{25\pi}{12},\;\tfrac{29\pi}{12}\;}$

Example 3 — $\cos\Bigl(\tfrac{1}{3}\theta\Bigr)= -\tfrac{\sqrt3}{2}$, restriction -2\pi\le\theta<0

• Reference angles for $\cos=-\tfrac{\sqrt3}{2}$: $\frac{5\pi}{6}$ and $\frac{7\pi}{6}$ (Q II & III).
• Build equations with the fractional angle:
  $\tfrac13\theta = \frac{5\pi}{6}+2k\pi, \qquad \tfrac13\theta = \frac{7\pi}{6}+2k\pi$
• Multiply every term by 3:
  $\theta = 5\pi + 6k\pi, \qquad \theta = 7\pi + 6k\pi$
• Test $k=-1,0,1$. All outputs are either greater than 0 or less than $-2\pi$, so no value lies in $[-2\pi,0)$.
• Result: No solution under the given restriction.

Example 4 — $\tan\bigl(\theta-\tfrac{\pi}{2}\bigr) = -1$, restriction 0\le\theta<2\pi

• $|\tan|=1$ reference angle: $\frac{\pi}{4}$.
• Negative result appears in Q II and Q IV ⇒ base angles $\frac{3\pi}{4},\;\frac{7\pi}{4}$.
• Equations (period $k\pi$ for tangent):
  $\theta-\frac{\pi}{2}=\frac{3\pi}{4}+k\pi, \qquad \theta-\frac{\pi}{2}=\frac{7\pi}{4}+k\pi$
• Add $\frac{\pi}{2}$ and put over a common denominator 4:
  $\theta = \frac{5\pi + 4k\pi}{4}, \qquad \theta = \frac{9\pi + 4k\pi}{4}$
• Evaluate $k=-1,0,1$:
  • $k=-1$ → $\theta=\frac{\pi}{4}$ and $\theta=\frac{5\pi}{4}$ — both inside.
  • $k=0$ → first formula repeats $\frac{\pi}{4}$, second gives \frac{9\pi}{4}>2\pi — reject.
  • $k=1$ → both exceed $2\pi$ — reject.
• Solutions: $\boxed{\;\tfrac{\pi}{4},\;\tfrac{5\pi}{4}\;}$

Example 5 — $\cot\bigl(3\theta - \tfrac{\pi}{6}\bigr) = \tfrac{\sqrt3}{3}$, restriction 0\le\theta<2\pi

• Recall $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$. A positive value implies same signs for $\cos$ and $\sin$ ⇒ quadrants I & III.
• $\cot=\frac{\sqrt3}{3}$ corresponds to $\theta_0=\frac{\pi}{3}$ (since $\cot\frac{\pi}{3}=\frac{1}{\sqrt3}$).
  Quadrant I: $\frac{\pi}{3}$; Quadrant III: $\frac{4\pi}{3}$.
• Set up equations with period $k\pi$:
  $3\theta-\frac{\pi}{6}=\frac{\pi}{3}+k\pi, \qquad 3\theta-\frac{\pi}{6}=\frac{4\pi}{3}+k\pi$
• Add $\frac{\pi}{6}$ then divide by 3:
  $\theta = \frac{\pi/3+\pi/6}{3}+\frac{k\pi}{3}=\frac{\pi}{6}+\frac{k\pi}{3}$
  $\theta = \frac{4\pi/3+\pi/6}{3}+\frac{k\pi}{3}=\frac{5\pi}{6}+\frac{k\pi}{3}$
• Write both over denominator 6:
  $\theta = \frac{\pi + 2k\pi}{6}, \qquad \theta = \frac{5\pi + 2k\pi}{6}$
• Evaluate $k=-1,0,1,2$ until $\theta\ge2\pi$:
  • $k=-1$ → negative first value, discard; second value $=\frac{\pi}{6}$ ✔
  • $k=0$ → $\frac{\pi}{6}$ (already listed) and $\frac{5\pi}{6}$ ✔
  • $k=1$ → $\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$ ✔ , second gives $\frac{7\pi}{6}$ ✔
  • $k=2$ begins to exceed $2\pi$.
• Accepted solutions (no duplicates):
  $\boxed{\;\tfrac{\pi}{6},\;\tfrac{\pi}{2},\;\tfrac{5\pi}{6},\;\tfrac{7\pi}{6}\;}$

Algebra & Notation Tips

• Common denominators: convert every fractional angle to the same denominator early; makes later additions/subtractions quick.
• Short-cut for stopping $k$:
  If $k=1$ already overshoots the interval, $k\ge2$ will overshoot even more.
• Sign tests for picking quadrants:
  • $\sin$ positive in Q I & II, negative in Q III & IV.
  • $\cos$ positive in Q I & IV, negative in Q II & III.
  • $\tan$ and $\cot$ positive in Q I & III, negative in Q II & IV.
• Typical mistake: forgetting to multiply the period by the coefficient in front of $\theta$ (e.g.
  $\sin(2\theta)$ has period $\pi$ for the argument but contributes $2k\pi$ once you solve for $\theta$).

Frequently Asked Questions Addressed in Class

• “Which $k$’s do we test?” → Always start with $k=-1,0,1,2$; extend only if your interval spans more than one revolution.
• “Is the restriction always $0!\to!2\pi$?” → Usually yes, but the book can give any one-revolution slice (e.g.
  $2\pi!\to!4\pi$ or $-2\pi!\to!0$). Read each problem.
• “Why cross out negative answers?” → If the lower bound of the interval is $0$, every negative result is automatically out.
• “What if the period is $\pi$?” → For $\tan$ & $\cot$ just add $k\pi$, not $2k\pi$.

Ending Reminders from the Instructor

• Homework matters: unsubmitted assignments cannot be retroactively added at semester’s end.
• A short practice set is posted on B12 covering these exact techniques; solutions are included for self-check.
• Instructor’s closing note: “Remember my sweet friend, Jesus. Once He’s inside your heart.”