Solving Trigonometric Equations

📐 Solving Trigonometric Equations

Part 1 of 7 — The Toolkit: Reference Angles & the Unit Circle

Topics in This Part

🔑 Key Concept: A trigonometric equation asks "for what angles is this true?" Because sine, cosine, and tangent repeat, most trig equations have infinitely many solutions. Before we solve, we need the unit-circle values cold — that's Part 1.

What Makes a Trig Equation Different

An equation like $x^2 = 4$ has two solutions, $x = 2$ and $x = -2$. Done.

A trig equation like $\sin\theta = \dfrac{1}{2}$ is different. Sine takes the value $\dfrac{1}{2}$ at $\theta = \dfrac{\pi}{6}$, but also at $\theta = \dfrac{5\pi}{6}$, and again at $\dfrac{\pi}{6} + 2\pi$, $\dfrac{5\pi}{6} + 2\pi$, and so on forever.

So solving a trig equation usually means three layers of work:

1. Find the reference angle — the acute angle whose trig value matches.
2. Place it in the right quadrant(s) — where is the function the correct sign?
3. Add the period to capture all solutions (Part 3).

💡 Almost every trig equation reduces to a unit-circle fact plus careful bookkeeping about quadrants and periods. Master the circle and the rest is logic.

Special Angles You Must Know

These are the values that show up on nearly every AP Precalculus problem. Memorize them in radians.

🔑 Memory aid: For $\sin$ of $0, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}$ write $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$. Cosine is the same list reversed.

Concept Check 🎯

Sign by Quadrant: "All Students Take Calculus"

The reference angle is the acute angle between the terminal side and the $x$-axis. It tells you the size of the trig value; the quadrant tells you the sign.

The mnemonic A·S·T·C ("All Students Take Calculus") walks quadrants I → II → III → IV.

⚠️ A common error: knowing $\sin\theta = \dfrac{1}{2}$ gives a reference angle of $\dfrac{\pi}{6}$, but sine is also positive in Quadrant II, so $\dfrac{5\pi}{6}$ is a solution too. Always ask: in which quadrants does the function have this sign?

Sign Detective 🔽

Use A·S·T·C to decide the sign of each value.

Reference Angles 🧮

The reference angle is the acute angle to the nearest $x$-axis. Write each reference angle as a multiple of $\pi$ and enter the coefficient (type 1/6 for $\frac{\pi}{6}$).

1) A terminal angle of $\frac{5\pi}{6}$ → reference angle $= \,?\cdot\pi$ 2) A terminal angle of $\frac{7\pi}{6}$ → reference angle $= \,?\cdot\pi$ 3) A terminal angle of $\frac{5\pi}{3}$ → reference angle $= \,?\cdot\pi$

📐 Solving Trigonometric Equations

Part 2 of 7 — Basic Equations on $[0, 2\pi)$

🔑 The Plan: To solve $\sin\theta = k$ (or $\cos$, or $\tan$) on one full turn $[0, 2\pi)$: find the reference angle from the unit circle, then use A·S·T·C to place a solution in each quadrant where the sign is correct. The same solutions are the boundaries for trig inequalities.

Solving a Sine Equation

Example: $\sin\theta = \dfrac{\sqrt{3}}{2}$ on $[0, 2\pi)$

1. Reference angle: $\sin^{-1}\dfrac{\sqrt{3}}{2} = \dfrac{\pi}{3}$.
2. Sign / quadrants: $\dfrac{\sqrt3}{2}$ is positive, and sine is positive in QI and QII.
3. Place it:
   • QI: $\theta = \dfrac{\pi}{3}$
   • QII: $\theta = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$

$\boxed{\theta = \frac{\pi}{3}, \; \frac{2\pi}{3}}$

✅ Check: $\sin\dfrac{2\pi}{3} = \sin\dfrac{\pi}{3} = \dfrac{\sqrt3}{2}$ ✓ (QII reference angle is $\pi - \theta$).

Solving a Cosine Equation (negative value)

Example: $\cos\theta = -\dfrac{1}{2}$ on $[0, 2\pi)$

1. Reference angle: $\cos^{-1}\dfrac{1}{2} = \dfrac{\pi}{3}$ (use the positive value for the reference angle).
2. Sign / quadrants: the value is negative, and cosine is negative in QII and QIII.
3. Place it:
   • QII: $\theta = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}$
   • QIII: $\theta = \pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$

$\boxed{\theta = \frac{2\pi}{3}, \; \frac{4\pi}{3}}$

💡 Quadrant formulas (reference angle $\alpha$): QI $= \alpha$, QII $= \pi - \alpha$, QIII $= \pi + \alpha$, QIV $= 2\pi - \alpha$.

Build the Solutions 🔽

You're solving $\cos\theta = -\dfrac{1}{2}$ with reference angle $\alpha = \dfrac{\pi}{3}$.

Solving a Tangent Equation

Tangent has period $\pi$ (not $2\pi$), so on $[0, 2\pi)$ a tangent equation usually has two solutions, exactly $\pi$ apart.

Example: $\tan\theta = 1$ on $[0, 2\pi)$

1. Reference angle: $\tan^{-1}(1) = \dfrac{\pi}{4}$.
2. Sign / quadrants: the value is positive, and tangent is positive in QI and QIII.
3. Place it: QI: $\dfrac{\pi}{4}$; QIII: $\pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}$.

$\boxed{\theta = \frac{\pi}{4}, \; \frac{5\pi}{4}}$

💡 Notice the two solutions differ by exactly $\pi$ — that's the signature of tangent.

From Equations to Inequalities

To solve an inequality such as $\sin\theta \ge k$ on $[0, 2\pi)$, solve the equation first. Its solutions are the boundary angles that split $[0, 2\pi)$ into intervals.

1. Boundaries: solve $\sin\theta = k$.
2. Test: pick one angle inside each interval and check the inequality (or read the unit circle: $\sin\theta$ is the $y$-coordinate, $\cos\theta$ is the $x$-coordinate).
3. Write the answer: keep the intervals that work. Include the boundary angles only for $\le$ or $\ge$.

Example 1: $\sin\theta \ge \dfrac{1}{2}$ on $[0, 2\pi)$

Boundaries: $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$. Test $\dfrac{\pi}{2}$: $\sin\dfrac{\pi}{2} = 1 \ge \dfrac12$ ✓. Test $\pi$: $\sin\pi = 0$ ✗.

$\boxed{\frac{\pi}{6} \le \theta \le \frac{5\pi}{6}}$

Example 2: $\cos\theta < -\dfrac{1}{2}$ on $[0, 2\pi)$

Boundaries: $\dfrac{2\pi}{3}$ and $\dfrac{4\pi}{3}$. Test $\pi$: $\cos\pi = -1 < -\dfrac12$ ✓. The inequality is strict, so the boundaries are excluded.

$\boxed{\frac{2\pi}{3} < \theta < \frac{4\pi}{3}}$

⚠️ Tangent: the angles where $\tan\theta$ is undefined ($\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$) also split the circle. For $\tan\theta > 1$ on $[0, 2\pi)$ the answer is $\dfrac{\pi}{4} < \theta < \dfrac{\pi}{2}$ or $\dfrac{5\pi}{4} < \theta < \dfrac{3\pi}{2}$.

Concept Check 🎯

Count & Place 🧮

Solve $\cos\theta = \dfrac{\sqrt{2}}{2}$ on $[0, 2\pi)$. Give each solution as a multiple of $\pi$ — enter the coefficient (e.g. type 1/4 for $\tfrac{\pi}{4}$).

1) Quadrant I solution: $\theta = \,?\,\cdot\pi$ 2) Quadrant IV solution: $\theta = \,?\,\cdot\pi$

📐 Solving Trigonometric Equations

Part 3 of 7 — All Solutions: The General Solution

🔑 The Big Idea: Because trig functions repeat, you capture every solution by adding whole periods. Sine and cosine repeat every $2\pi$; tangent repeats every $\pi$. We write "$+ 2\pi k$" or "$+ \pi k$" where $k$ is any integer.

Adding the Period

Once you have the solutions on $[0, 2\pi)$, the general solution adds one full period times an integer $k = \ldots, -2, -1, 0, 1, 2, \ldots$

Sine / Cosine (period $2\pi$)

If $\theta = \dfrac{\pi}{6}$ and $\theta = \dfrac{5\pi}{6}$ solve $\sin\theta = \dfrac12$ on one turn, then all solutions are:

$\theta = \frac{\pi}{6} + 2\pi k \quad \text{or} \quad \theta = \frac{5\pi}{6} + 2\pi k, \quad k \in \mathbb{Z}$

Tangent (period $\pi$)

Because tangent's two solutions per turn already differ by $\pi$, you only need one base solution:

$\tan\theta = 1 \;\Rightarrow\; \theta = \frac{\pi}{4} + \pi k, \quad k \in \mathbb{Z}$

💡 One $+\pi k$ family for tangent automatically generates both the QI and QIII answers.

Concept Check 🎯

Listing Solutions on a Given Interval

Often a problem restricts the interval (e.g. $[0, 2\pi)$ or $[0, 4\pi)$). The trick: write the general solution, then plug in integers $k = 0, 1, 2, \ldots$ and keep only the angles inside the interval.

Example: solve $\sin\theta = \dfrac12$ on $[0, 4\pi)$

General solution: $\theta = \dfrac{\pi}{6} + 2\pi k$ or $\dfrac{5\pi}{6} + 2\pi k$.

So on $[0, 4\pi)$ there are four solutions: $\dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{13\pi}{6}, \dfrac{17\pi}{6}$.

⚠️ Note $4\pi = \dfrac{24\pi}{6}$, so $\dfrac{25\pi}{6}$ is just past the right edge and is excluded.

General-Solution Builder 🔽

You found $\theta = \dfrac{\pi}{3}$ and $\theta = \dfrac{5\pi}{3}$ solve $\cos\theta = \dfrac12$ on one turn.

Count the Solutions 🧮

How many solutions does each equation have on the interval $[0, 2\pi)$? (Enter a whole number.)

1) $\cos\theta = 0$ 2) $\sin\theta = 1$ 3) $\tan\theta = \sqrt{3}$

📐 Solving Trigonometric Equations

Part 4 of 7 — Equations That Need Algebra First

🔑 The Move: Many trig equations aren't "isolated" yet. Treat the trig function like a single variable — call it $u$ — and use ordinary algebra (isolate, factor, quadratic formula) to get $\sin\theta = \text{number}$, then solve with the unit circle.

Step 1: Isolate the Trig Function

Example: $2\sin\theta - 1 = 0$ on $[0, 2\pi)$

Solve for the trig piece exactly like a linear equation:

$2\sin\theta = 1 \;\Rightarrow\; \sin\theta = \frac{1}{2}$

Now it's a basic equation: reference angle $\dfrac{\pi}{6}$, sine positive in QI and QII:

$\boxed{\theta = \frac{\pi}{6}, \; \frac{5\pi}{6}}$

💡 Get the equation into the form $(\text{trig}) = (\text{number})$ first. Everything you know from Parts 2–3 then applies.

Step 2: Factoring (the zero-product idea)

If two factors multiply to $0$, at least one factor is $0$.

Example: $\sin\theta\,(2\cos\theta + 1) = 0$ on $[0, 2\pi)$

Set each factor to $0$:

Factor A: $\sin\theta = 0 \;\Rightarrow\; \theta = 0, \, \pi$

Factor B: $2\cos\theta + 1 = 0 \Rightarrow \cos\theta = -\dfrac{1}{2} \Rightarrow \theta = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}$

$\boxed{\theta = 0, \; \frac{2\pi}{3}, \; \pi, \; \frac{4\pi}{3}}$

⚠️ Never divide both sides by $\sin\theta$ — you would lose the $\sin\theta = 0$ solutions. Factor instead and set each piece to zero.

Concept Check 🎯

Step 3: Quadratic-Form Equations

When an equation is quadratic in a trig function, substitute $u = \sin\theta$ (or $\cos\theta$), factor, then back-substitute.

Example: $2\sin^2\theta - \sin\theta - 1 = 0$ on $[0, 2\pi)$

Let $u = \sin\theta$: $\;2u^2 - u - 1 = 0$. Factor: $(2u + 1)(u - 1) = 0$.

• $u = 1 \Rightarrow \sin\theta = 1 \Rightarrow \theta = \dfrac{\pi}{2}$
• $u = -\dfrac{1}{2} \Rightarrow \sin\theta = -\dfrac{1}{2} \Rightarrow \theta = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$

$\boxed{\theta = \frac{\pi}{2}, \; \frac{7\pi}{6}, \; \frac{11\pi}{6}}$

💡 $\sin\theta = -\dfrac12$ is negative → QIII and QIV: $\pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6}$ and $2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6}$.

Factor & Solve 🔽

Solve $2\cos^2\theta - \cos\theta = 0$ on $[0, 2\pi)$ by factoring.

Isolate, Then Solve 🧮

Solve $2\cos\theta + \sqrt{3} = 0$ on $[0, 2\pi)$. Give each solution as a coefficient of $\pi$ (type 5/6 for $\tfrac{5\pi}{6}$).

1) Smaller solution: $\theta = \,?\,\cdot\pi$ 2) Larger solution: $\theta = \,?\,\cdot\pi$

📐 Solving Trigonometric Equations

Part 5 of 7 — Inverse Trig & the Calculator

🔑 When values aren't "special": For $\cos\theta = 0.4$ there's no unit-circle answer. Use the inverse trig keys ($\sin^{-1}, \cos^{-1}, \tan^{-1}$) to get the reference angle, then place solutions in every correct quadrant yourself.

The Calculator Only Gives ONE Answer

Inverse trig functions are built to return a single angle from a restricted range:

So the calculator's answer is just one solution (or a reference angle). You must find the others using A·S·T·C and the period.

⚠️ Mode matters! Set your calculator to radians when the problem uses radians and degrees when it uses degrees. A wrong mode is the #1 source of wrong trig-equation answers.

Worked Example with a Calculator

Solve $\cos\theta = 0.4$ on $[0, 2\pi)$ (round to 2 decimals)

1. Reference angle: $\cos^{-1}(0.4) \approx 1.16$ rad. Since $0.4 > 0$, this is already a QI angle.
2. Quadrants: cosine is positive in QI and QIV.
3. Place it:
   • QI: $\theta \approx 1.16$
   • QIV: $\theta = 2\pi - 1.16 \approx 6.28 - 1.16 = 5.12$

$\boxed{\theta \approx 1.16, \; 5.12}$

💡 If the value were negative, $\cos^{-1}$ returns a QII angle directly; you'd then get the QIII partner from $2\pi - (\text{that angle})$.

Concept Check 🎯

From Calculator to All Solutions 🔽

You're solving $\sin\theta = 0.6$ on $[0, 2\pi)$ and your calculator (radian mode) gives $\sin^{-1}(0.6) \approx 0.64$.

Calculator Practice 🧮

Solve $\tan\theta = 2$ on $[0, 2\pi)$. Your calculator gives $\tan^{-1}(2) \approx 1.11$ rad. Round answers to 2 decimals.

1) Quadrant I solution: $\theta \approx \,?$ 2) Quadrant III solution: $\theta \approx \,?$ (add $\pi$)

📐 Solving Trigonometric Equations

Part 6 of 7 — Applications: Periodic Models

🔑 Why this matters: Tides, daylight hours, temperature, sound, and AC voltage all rise and fall periodically. They're modeled by $f(t) = A\sin\big(B(t - C)\big) + D$, and answering "when does it reach a certain value?" means solving a trig equation.

Reading a Sinusoidal Model

For $f(t) = A\sin\big(B(t - C)\big) + D$:

The maximum is $D + |A|$ and the minimum is $D - |A|$.

Example: a Ferris wheel

A rider's height is $h(t) = 25\sin\big(\tfrac{\pi}{15}(t - 7.5)\big) + 30$ meters, $t$ in seconds.

• Amplitude $25$, midline $30$ → height ranges from $5$ m to $55$ m.
• Period $= \dfrac{2\pi}{\pi/15} = 30$ seconds per revolution.

💡 To find when the rider is at a given height, set $h(t)$ equal to that height and solve the trig equation for $t$.

Concept Check 🎯

Solving an Applied Equation

When is the temperature $T(t) = 10\sin\big(\tfrac{\pi}{12}t\big) + 60$ equal to $65^\circ$? (first time, $t \ge 0$ hours)

1. Set up: $10\sin\big(\tfrac{\pi}{12}t\big) + 60 = 65$.
2. Isolate: $10\sin\big(\tfrac{\pi}{12}t\big) = 5 \Rightarrow \sin\big(\tfrac{\pi}{12}t\big) = \dfrac{1}{2}$.
3. Solve the inner angle: let $u = \tfrac{\pi}{12}t$. The first positive solution of $\sin u = \dfrac12$ is $u = \dfrac{\pi}{6}$.
4. Back-substitute: $\dfrac{\pi}{12}t = \dfrac{\pi}{6} \Rightarrow t = \dfrac{\pi}{6}\cdot\dfrac{12}{\pi} = 2$ hours.

$\boxed{t = 2 \text{ hours}}$

⚠️ Substitution is key: solve for the inside angle first ($u = \tfrac{\pi}{12}t$), then undo it to find $t$.

Inequalities in context: when is $T(t) > 65$ during the first day ($0 \le t < 24$)?

The context restricts the domain to one day. Then $\sin u > \dfrac12$ with $u = \tfrac{\pi}{12}t$ in $[0, 2\pi)$ means $\dfrac{\pi}{6} < u < \dfrac{5\pi}{6}$. Undo the inner angle: $\dfrac{\pi}{6}\cdot\dfrac{12}{\pi} = 2$ and $\dfrac{5\pi}{6}\cdot\dfrac{12}{\pi} = 10$, so the temperature is above $65$ for $2 < t < 10$ hours.

Set Up the Equation 🔽

The depth of water at a dock is $D(t) = 4\sin\big(\tfrac{\pi}{6}t\big) + 9$ feet, $t$ in hours. You want to know when the depth is $11$ feet.

Daylight Model 🧮

Hours of daylight are modeled by $L(t) = 3\sin\big(\tfrac{\pi}{6}t\big) + 12$, where $t$ is months after the spring equinox.

1) What is the maximum number of daylight hours? $L_{\max} = \,?$ 2) What is the period, in months? $\,?$ 3) Solve $\sin\big(\tfrac{\pi}{6}t\big) = 1$ for the smallest $t > 0$ (months to the longest day): $t = \,?$

📐 Solving Trigonometric Equations

Part 7 of 7 — Mixed Practice & Mastery Check

You can now (1) read the unit circle, (2) solve basic equations on $[0, 2\pi)$, (3) write general solutions, (4) handle algebra and factoring, (5) use inverse trig, and (6) solve applied periodic models. Let's pull it all together.

Quick Reference

⚠️ Top pitfalls: dropping a quadrant, dropping the $\pm$ / $+2\pi k$, dividing both sides by a trig factor (lose solutions), and leaving the calculator in the wrong angle mode.

Mixed Strategy 🔽

Match each equation to the first thing you should do.

Mixed Practice 🧮

1) Solve $2\sin\theta - \sqrt{3} = 0$ on $[0,2\pi)$. Smaller solution as a coefficient of $\pi$ (type 1/3 for $\tfrac{\pi}{3}$): $\theta = \,?\,\cdot\pi$ 2) How many solutions does $\cos\theta = -1$ have on $[0, 2\pi)$? 3) For $f(t) = 6\sin\big(\tfrac{\pi}{4}t\big) + 10$, find the period.

Exit Quiz ✅

Answer all three to finish the lesson.