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Solving Trig Equations
How to Solve Trig Equations: Linear, Quadratic, Substitution & Multiple Angle
Quick Answer — The Four Equation Types
Linear
2sin θ − 1 = 0 → isolate, use inverse trig
Quadratic
2sin²θ − 1 = 0 → factor or quadratic formula
Pythagorean substitution
Two functions → reduce to one using identities
Multiple angle
sin 2θ = c → expand identity, solve, more solutions
All four types end the same way: get trig function = value, find the reference angle, identify the correct quadrants, list all solutions in the given interval or write the general solution.
In This Guide
- How Trig Equations Work
- Which Technique to Use — Decision Tree
- Interval Solutions vs. General Solutions
- Type 1 — Linear Equations
- Type 2 — Quadratic Equations
- Type 3 — Pythagorean Substitution
- Type 4 — Multiple Angle Equations
- Phase-Shift Equations
- Common Mistakes
- Practice Problems
- Platform Tips
- Frequently Asked Questions
How Trig Equations Work
A trig equation is an equation that is true only at specific angles — unlike a trig identity, which is true for every angle. When you solve a trig equation, you’re finding the angles that make it true.
The complication that makes trig equations different from regular algebra is that every trig function is periodic. Sin θ = 0.5 is true at 30°, and also at 150°, and also at 30° + 360°, and at 150° + 360°, and so on forever. This means every trig equation either has infinitely many solutions (if you want all of them) or a finite list of solutions within a specified interval (if the problem gives you bounds like [0°, 360°] or [0, 2π]).
The process for every equation type is the same four steps: get the trig function isolated on one side, apply inverse trig to find the reference angle, use quadrant analysis to find all solutions in one period, then either list those solutions (interval) or add the period offset (general solution).
Which Technique to Use
Before solving anything, identify what type of equation you have. The four types require different opening moves, but they all converge on the same finishing process.
All four paths reduce to the same final step: trig function = value → reference angle → quadrant check → list solutions.
Interval Solutions vs. General Solutions
Most exam problems specify which format they want. Getting the right answers in the wrong format costs full marks.
The unit circle always produces exactly two solutions per period for sin and cos (one or zero for specific values), plus infinite repetitions spaced one period apart.
Interval Solution
The problem says “find all θ in [0°, 360°]” or “on [0, 2π].”
List only the angles inside that interval. For sin θ = 0.5: θ = 30° and θ = 150°.
No +2πk needed. Just the specific values.
General Solution
The problem says “find all solutions” with no interval restriction.
Write each solution with a period offset: θ = π/6 + 2πk and θ = 5π/6 + 2πk, k ∈ ℤ.
Use +2πk for sin/cos/sec/csc. Use +πk for tan/cot (period = π).
⚠️ Tangent General Solution Uses Only One Expression, Not Two
For sin and cos, the general solution always requires two separate expressions — one for each quadrant where the function has the correct sign. For example, sin θ = 1/2 gives θ = π/6 + 2πk and θ = 5π/6 + 2πk.
For tan, only one expression is needed: θ = α + πk. Because the period of tangent is π, adding πk to a Q I solution automatically generates the Q III solution (and every other repetition). Writing two separate expressions for a tangent general solution is a common error — it produces the right values but doubles the apparent solution count, which platforms flag as incorrect.
Type 1 — Linear Equations
The most common type. One trig function, no squares, just constants and coefficients. The approach is identical to solving a linear algebra equation: isolate the trig function, then apply inverse trig to find the reference angle.
Isolate the trig function on one side of the equation.
Apply the appropriate inverse function to find the reference angle (always positive, always in Q I).
Determine which quadrants give the correct sign for the function (ASTC). A negative value means the function is negative — find the two quadrants where it’s negative.
Build the solutions: Q I = ref, Q II = 180° − ref, Q III = 180° + ref, Q IV = 360° − ref. Keep only those in the requested interval.
Keep this open while working through problems. The radian and degree formulas are equivalent — use whichever matches the interval your problem specifies.
📐 Worked Example — Solve: 2sin θ − 1 = 0 on [0°, 360°]
Step 1 — Isolate: 2sin θ = 1 → sin θ = 1/2
Step 2 — Reference angle: arcsin(1/2) = 30°
Step 3 — Quadrants: sin is positive in Q I and Q II.
Step 4 — Solutions: Q I: θ = 30°. Q II: θ = 180° − 30° = 150°.
Answer: θ = 30° and θ = 150°
📐 Worked Example — Solve: √3 tan θ + 1 = 0 on [0, 2π]
Step 1 — Isolate: tan θ = −1/√3 = −√3/3
Step 2 — Reference angle: arctan(√3/3) = 30° = π/6
Step 3 — Quadrants: tan is negative in Q II and Q IV.
Step 4 — Solutions: Q II: θ = π − π/6 = 5π/6. Q IV: θ = 2π − π/6 = 11π/6.
Answer: θ = 5π/6 and θ = 11π/6
✅ Always Verify Your Solutions
Substitute each answer back into the original equation to confirm it works. For 2sin θ − 1 = 0 with θ = 30°: 2sin(30°) − 1 = 2(0.5) − 1 = 0 ✓. With θ = 150°: 2sin(150°) − 1 = 2(0.5) − 1 = 0 ✓.
This takes 30 seconds and catches sign errors, wrong quadrant choices, and mode errors (degrees vs radians) before they cost marks on ALEKS or Hawkes.
Type 2 — Quadratic Equations
When a trig function appears squared, treat the entire squared expression as a single variable and apply quadratic methods. Let u = sin θ (or cos θ, or tan θ), substitute, solve the quadratic for u, then solve each result as a linear trig equation.
⚠️ Check Your Solutions Against the Domain
After factoring, each factor gives a value for the trig function. Before applying inverse trig, check that the value is in the valid range: sin and cos must be between −1 and 1. If a factor gives sin θ = 3 or cos θ = −2, that factor has no real solutions — discard it.
📐 Worked Example — Solve: 2sin²θ − sin θ − 1 = 0 on [0°, 360°]
Step 1 — Recognize as quadratic: Let u = sin θ → 2u² − u − 1 = 0
Step 2 — Factor: (2u + 1)(u − 1) = 0 → u = −1/2 or u = 1
Step 3 — Solve each case:
Case A: sin θ = −1/2. Reference angle = 30°. Sin is negative in Q III and Q IV. → θ = 210° and θ = 330°
Case B: sin θ = 1. → θ = 90° (only one solution — sin = 1 only at 90°)
Answer: θ = 90°, 210°, 330°
📐 Worked Example — Solve: 2cos²θ − 1 = 0 on [0, 2π]
Step 1 — Isolate: cos²θ = 1/2 → cos θ = ±1/√2 = ±√2/2
Step 2 — Two cases:
Case A: cos θ = √2/2 → reference angle π/4, cos positive in Q I and Q IV → θ = π/4 and θ = 7π/4
Case B: cos θ = −√2/2 → same reference angle π/4, cos negative in Q II and Q III → θ = 3π/4 and θ = 5π/4
Answer: θ = π/4, 3π/4, 5π/4, 7π/4
Type 3 — Pythagorean Substitution
When an equation contains two different trig functions (like sin and cos together, or tan and sec), it can’t be solved directly. The goal is to eliminate one function using a Pythagorean identity so the equation contains only one trig function — then it becomes either linear or quadratic.
The substitution to reach for depends on which two functions appear. If you see sin and cos together with a squared term, replace the squared one using cos²θ = 1 − sin²θ or sin²θ = 1 − cos²θ. If you see tan and sec together, use sec²θ = tan²θ + 1. The aim is always to get a single function throughout.
📐 Worked Example — Solve: cos²θ + sin θ + 1 = 0 on [0°, 360°]
Step 1 — Identify: Two functions (sin and cos). Replace cos²θ.
Step 2 — Substitute: (1 − sin²θ) + sin θ + 1 = 0 → −sin²θ + sin θ + 2 = 0
Step 3 — Rearrange: sin²θ − sin θ − 2 = 0
Step 4 — Factor: (sin θ − 2)(sin θ + 1) = 0 → sin θ = 2 (impossible) or sin θ = −1
Step 5 — Solve: sin θ = −1 → θ = 270°
Answer: θ = 270°
📐 Worked Example — Solve: sec²θ − 2tan θ − 4 = 0 on [0°, 360°]
Step 1 — Identify: sec and tan together. Use sec²θ = tan²θ + 1.
Step 2 — Substitute: (tan²θ + 1) − 2tan θ − 4 = 0 → tan²θ − 2tan θ − 3 = 0
Step 3 — Factor: (tan θ − 3)(tan θ + 1) = 0 → tan θ = 3 or tan θ = −1
Step 4 — Solve each case:
Case A: tan θ = 3 → reference angle = arctan(3) ≈ 71.6°. Tan positive in Q I and Q III → θ ≈ 71.6° and θ ≈ 251.6°
Case B: tan θ = −1 → reference angle = 45°. Tan negative in Q II and Q IV → θ = 135° and θ = 315°
Answer: θ ≈ 71.6°, 135°, 251.6°, 315°
Type 4 — Multiple Angle Equations
When the argument of the trig function is a multiple of θ — like sin 2θ, cos 3θ, or tan(θ/2) — the equation has more solutions in any given interval than a standard equation would. The reason: if you’re solving for 2θ in [0°, 360°], then 2θ ranges over [0°, 720°] — a full two periods — so you get twice as many solutions.
The cleanest approach is substitution: let u = 2θ (or 3θ, or θ/2), solve for u in the expanded interval, then divide every solution by 2 (or 3, or multiply by 2) to recover θ.
📐 Worked Example — Solve: sin 2θ = √3/2 on [0°, 360°]
Step 1 — Substitute: Let u = 2θ. Since θ ∈ [0°, 360°], u ∈ [0°, 720°].
Step 2 — Solve sin u = √3/2: Reference angle = 60°. Sin positive in Q I and Q II.
Step 3 — List all u in [0°, 720°]:
First period: u = 60° and u = 120°
Second period: u = 60° + 360° = 420° and u = 120° + 360° = 480°
Step 4 — Recover θ: θ = u/2 → θ = 30°, 60°, 210°, 240°
Answer: θ = 30°, 60°, 210°, 240°
📐 Worked Example — Solve: cos(θ/2) = −1 on [0°, 360°]
Step 1 — Substitute: Let u = θ/2. Since θ ∈ [0°, 360°], u ∈ [0°, 180°].
Step 2 — Solve cos u = −1 in [0°, 180°]: cos u = −1 only at u = 180°.
Step 3 — Recover θ: θ = 2u = 2 × 180° = 360°.
Note: 360° is the boundary of [0°, 360°]. Whether to include it depends on whether the interval is open or closed — here it’s closed, so include it.
Answer: θ = 360°
Phase-Shift Equations
Phase-shift equations look like Type 4 but with an added constant inside the argument: sin(θ + π/4) = c, cos(2θ − π/3) = c, tan(θ − π/6) = c. They appear regularly on ALEKS and Hawkes as a natural extension of multiple-angle problems. The technique is the same — substitute, solve for the inner expression, then recover θ.
The key difference from plain multiple-angle equations: after solving for the inner expression, recovering θ requires subtracting the phase shift (or adding it, depending on the sign). This is where most errors occur — students divide correctly but forget to adjust for the shift.
Let u equal the entire inner expression: u = 2θ − π/3, u = θ + π/4, etc.
Find the interval for u. If θ ∈ [0, 2π] and u = 2θ − π/3, then u ∈ [−π/3, 2π·2 − π/3] = [−π/3, 11π/3]. Adjust for any negative lower bound by including one extra period below zero.
Solve trig(u) = c in the expanded interval for u, using the standard quadrant method.
Recover θ from each solution: rearrange u = 2θ − π/3 to θ = (u + π/3)/2. Discard any θ outside the original interval.
📐 Worked Example — Solve: sin(2θ − π/3) = 1/2 on [0, 2π]
Step 1 — Substitute: Let u = 2θ − π/3.
Step 2 — Interval for u: θ ∈ [0, 2π] → u = 2θ − π/3 ∈ [−π/3, 4π − π/3] = [−π/3, 11π/3].
Step 3 — Solve sin u = 1/2 in [−π/3, 11π/3]: Reference angle = π/6. Sin positive in Q I and Q II.
Period 1 (starting from just below 0): u = π/6 and u = 5π/6
Period 2: u = π/6 + 2π = 13π/6 and u = 5π/6 + 2π = 17π/6
Check: −π/3 ≈ −1.05, 11π/3 ≈ 11.52. All four values (π/6≈0.52, 5π/6≈2.62, 13π/6≈6.81, 17π/6≈8.90) fall in range ✓
Step 4 — Recover θ: θ = (u + π/3)/2
θ = (π/6 + π/3)/2 = (π/2)/2 = π/4
θ = (5π/6 + π/3)/2 = (7π/6)/2 = 7π/12
θ = (13π/6 + π/3)/2 = (15π/6)/2 = 5π/4
θ = (17π/6 + π/3)/2 = (19π/6)/2 = 19π/12
Answer: θ = π/4, 7π/12, 5π/4, 19π/12
Common Mistakes
❌ Finding only one solution when there are two
arcsin(0.5) gives 30° — one answer. But sin θ = 0.5 has two solutions per period: 30° and 150°. Students who stop after finding the inverse trig result miss the second solution in the other quadrant. Always check both quadrants where the function has the correct sign.
❌ Forgetting to double the interval for multiple angle equations
For sin 2θ = c on [0°, 360°], you must find solutions for u = 2θ in [0°, 720°] — not [0°, 360°]. Solving in the wrong interval misses the second set of solutions entirely. Always multiply the interval bounds by the same coefficient as θ.
❌ Dividing by a trig function instead of factoring
If you have sin θ · cos θ = sin θ, don’t divide both sides by sin θ — that cancels solutions where sin θ = 0. Instead, move everything to one side: sin θ · cos θ − sin θ = 0, factor out sin θ: sin θ(cos θ − 1) = 0, then set each factor to zero. Dividing by a trig function always risks losing solutions.
❌ Using the wrong period for the general solution
Sin, cos, sec, and csc repeat every 2π — add +2πk. Tan and cot repeat every π — add +πk. Writing +2πk for a tangent equation gives the right structure but misses every other solution family. Check which function you’re working with before writing the general solution.
❌ Keeping invalid solutions from quadratic equations
After factoring a quadratic trig equation, each factor produces a value for the trig function. Always check that value is in the valid range before proceeding. sin θ = 2 and cos θ = −1.5 have no solutions. Keeping them and trying to apply inverse trig will give a calculator error or a nonsense answer.
❌ Using degrees when the problem expects radians (or vice versa)
If the interval is given as [0, 2π], your answers must be in radians. If it’s given as [0°, 360°], use degrees. Platforms like ALEKS and WebAssign will mark the correct numerical value wrong if it’s in the wrong unit. When in doubt about which mode your calculator is in, verify with a known value: sin(30°) = sin(π/6) = 0.5.
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Practice Problems
📝 Problem 1 — Linear
Solve: 2cos θ + √3 = 0 on [0°, 360°]
Click to reveal answer
Isolate: cos θ = −√3/2
Reference angle: arccos(√3/2) = 30°
cos is negative in Q II and Q III.
Answer: θ = 150° and θ = 210°
📝 Problem 2 — Quadratic
Solve: 2cos²θ + cos θ = 0 on [0, 2π]
Click to reveal answer
Factor: cos θ(2cos θ + 1) = 0
Case A: cos θ = 0 → θ = π/2 and θ = 3π/2
Case B: cos θ = −1/2 → reference angle π/3, cos negative in Q II and Q III → θ = 2π/3 and θ = 4π/3
Answer: θ = π/2, 2π/3, 4π/3, 3π/2
📝 Problem 3 — Pythagorean Substitution
Solve: sin²θ − cos θ − 1 = 0 on [0°, 360°]
Click to reveal answer
Replace sin²θ with 1 − cos²θ: (1 − cos²θ) − cos θ − 1 = 0 → −cos²θ − cos θ = 0
Multiply by −1: cos²θ + cos θ = 0
Factor: cos θ(cos θ + 1) = 0
Case A: cos θ = 0 → θ = 90° and θ = 270°
Case B: cos θ = −1 → θ = 180°
Answer: θ = 90°, 180°, 270°
📝 Problem 4 — Multiple Angle
Solve: cos 2θ = 1/2 on [0°, 360°]
Click to reveal answer
Let u = 2θ. θ ∈ [0°, 360°] so u ∈ [0°, 720°].
Solve cos u = 1/2: reference angle = 60°. Cos positive in Q I and Q IV.
First period: u = 60° and u = 300°
Second period: u = 420° and u = 660°
Recover θ = u/2: θ = 30°, 150°, 210°, 330°
Answer: θ = 30°, 150°, 210°, 330°
Platform Tips
ALEKS — interval solutions on ALEKS must be entered as a list separated by commas. Entering θ = 30°, 150° as a single expression with “and” between them will be rejected. ALEKS also uses radians by default in many trig modules — check the problem statement carefully. If the interval is written as [0, 2π], enter radian answers only.
Hawkes Learning — Certify mode for trig equations requires exact answers, not decimal approximations. arctan(3) ≈ 71.6° is acceptable only if the problem explicitly asks for decimal. Otherwise Hawkes expects exact values from the standard angle table (30°, 45°, 60°, 90°, and their quadrant equivalents). Problems with non-standard angles will specify “round to the nearest tenth.”
WebAssign — general solution problems on WebAssign require the +2πk notation exactly as written. Some problems accept n instead of k, some use n as the variable — read the problem statement to match the expected variable name. Entering θ = π/6 + 2πn when the problem expects θ = π/6 + 2πk may or may not be marked correct depending on the instructor’s settings.
Knewton Alta — adaptive trig equation problems escalate from linear to quadratic to substitution types as you demonstrate mastery. If the platform keeps presenting quadratic or substitution problems, it means you’ve already passed the linear threshold. Work through the substitution examples on this page before the session to avoid getting stuck mid-problem.
MyMathLab — the equation editor for trig answers requires using the “pi” button for π rather than typing the letter p. Entering 3.14159 for π will be marked wrong. For general solutions, the +2nπ entry requires the multiplication symbol from the editor — don’t type 2nπ as plain text without the explicit operator between 2, n, and π.
Frequently Asked Questions
Why does sin θ = 0.5 have two solutions instead of one?
Because sine is not one-to-one over a full period. The horizontal line y = 0.5 intersects the sine curve at two points in every 360° interval — once on the way up (at 30°) and once on the way down (at 150°). This is why we check both quadrants where sine is positive, not just the one arcsin returns.
How do I know when to use +2πk vs +πk in the general solution?
Match the period of the function. Sin, cos, sec, and csc all have period 2π — use +2πk. Tan and cot have period π — use +πk. If you’re solving a multiple-angle equation like sin 2θ = c, the general solution is in terms of the substituted variable u = 2θ, so you write u = value + 2πk, then divide through by 2 to recover θ = value/2 + πk.
What if a trig equation has no solution?
It happens when the equation requires a trig function to take a value outside its range. Sin and cos are bounded between −1 and 1, so sin θ = 2 has no solution. Tan and cot are unbounded, so they always have solutions for any real value. When you factor a quadratic trig equation and one factor gives an impossible value, discard it and continue with the other factor only.
Can I just use a calculator to solve trig equations?
For finding the reference angle, yes. But a calculator only gives you one answer — the principal value from the inverse trig function — and most equations have two solutions per period. You still have to use quadrant analysis to find the second solution, and you still have to handle the period offset for general solutions or multiple-angle equations. The calculator handles the arithmetic; the method handles the reasoning.
What’s the difference between a trig equation and a trig identity?
A trig identity is true for all values of θ — sin²θ + cos²θ = 1 holds at every angle. A trig equation is only true at specific angles — 2sin θ − 1 = 0 is only true at θ = 30°, 150°, and their coterminal angles. Solving a trig equation means finding those specific angles. Proving a trig identity means showing it holds everywhere, not finding where it holds.
How many solutions should I expect in [0°, 360°]?
For a linear trig equation: typically 2 solutions (one per quadrant where the function has the right sign), sometimes 1 (at exactly 0°, 90°, 180°, or 270°), occasionally 0 (impossible value). For a quadratic with two factors: up to 4 solutions. For sin 2θ = c on [0°, 360°]: typically 4 solutions (two per period, two periods). Always verify your solutions by substituting back into the original equation if unsure.
Need Help With Trig Equations?
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Related Resources
Prerequisites:
- Inverse Trig Functions — needed to find reference angles from trig values
- Pythagorean Identities — needed for Type 3 substitution equations
- Unit Circle Explained — needed for quadrant analysis and exact values
What comes next:
- Sum and Difference Formulas — expands the set of solvable equations
- Double and Half Angle Formulas — essential for Type 4 multiple-angle equations
- Do My Trigonometry Homework