What Are Trigonometric Equations?
A trigonometric equation is an equation that contains one or more trigonometric functions of an unknown angle. Unlike a trigonometric identity (which is true for every value of the variable), a trigonometric equation is only satisfied by specific angle values. Your job when solving a trig equation is to find all of those values.
For example, the equation sin θ = 1/2 is not true for every angle. It is only true when θ equals 30° (or π/6 radians), 150° (or 5π/6 radians), and any angle coterminal with those. Solving trigonometric equations requires a combination of algebraic techniques (factoring, substitution, isolating variables) and knowledge of trigonometric values and identities.
This lesson covers the core methods you will use again and again: solving basic equations of the form sin x = a, finding general solutions, factoring, applying identities, handling quadratic forms, and combining multiple strategies for complex problems.
Basic Trig Equations: sin x = a, cos x = a, tan x = a
The simplest trigonometric equations involve a single trig function set equal to a constant. The approach is always the same: isolate the trig function, then determine which angles produce that value using known reference angles and the unit circle.
Solving sin x = a
When solving sin x = a (where −1 ≤ a ≤ 1), first find the reference angle α = arcsin(|a|). Then determine which quadrants sine has the correct sign:
- If a > 0: solutions lie in Quadrant I (x = α) and Quadrant II (x = π − α)
- If a < 0: solutions lie in Quadrant III (x = π + α) and Quadrant IV (x = 2π − α)
- If a = 0: x = 0 and x = π
Solving cos x = a
For cos x = a (where −1 ≤ a ≤ 1), find the reference angle α = arccos(|a|). Then:
- If a > 0: solutions lie in Quadrant I (x = α) and Quadrant IV (x = 2π − α)
- If a < 0: solutions lie in Quadrant II (x = π − α) and Quadrant III (x = π + α)
Solving tan x = a
For tan x = a (any real value of a), find the reference angle α = arctan(|a|). Since tangent is positive in Quadrants I and III, and negative in Quadrants II and IV:
- If a > 0: x = α (QI) and x = π + α (QIII)
- If a < 0: x = π − α (QII) and x = 2π − α (QIV)
Example 1: Solve sin x = √3/2 on [0, 2π)
Step 1: Find the reference angle. We know sin(π/3) = √3/2, so the reference angle is α = π/3.
Step 2: Since √3/2 > 0, sine is positive in Quadrants I and II.
Step 3: Quadrant I solution: x = π/3.
Step 4: Quadrant II solution: x = π − π/3 = 2π/3.
Answer: x = π/3 and x = 2π/3.
Key Takeaway
For basic trig equations, always start by finding the reference angle, then use the sign of the constant and the ASTC rule (All Students Take Calculus) to determine which quadrants contain solutions.
Finding All Solutions: General Solutions
Trigonometric functions are periodic, meaning they repeat their values at regular intervals. When a problem asks you to find all solutions (not just those in a specific interval), you must express the answer as a general solution that accounts for every full period of repetition.
General Solutions for Sine and Cosine
Since sine and cosine have a period of 2π, if x₀ is one solution, then x₀ + 2nπ is also a solution for any integer n. Because each basic equation has two solutions per period, the general solution includes both families:
General Solution for Tangent
Tangent has a period of π, and it produces only one solution per period. This makes its general solution simpler:
Example 2: Find All Solutions of cos x = −1/2
Step 1: Find the reference angle. cos(π/3) = 1/2, so α = π/3.
Step 2: Since the value is negative, cosine is negative in Quadrants II and III. The two base solutions on [0, 2π) are x = π − π/3 = 2π/3 and x = π + π/3 = 4π/3.
Step 3: Write the general solution by adding 2nπ to each:
x = 2π/3 + 2nπ or x = 4π/3 + 2nπ, where n is any integer.
Factoring Trig Equations
Many trigonometric equations can be solved by factoring, just as you would factor a polynomial equation. The key principle is the same: if a product of factors equals zero, then at least one of the factors must be zero. This is the zero-product property.
Typical situations where factoring applies include equations with a common trig factor that can be pulled out, and equations that resemble factorable polynomial patterns such as difference of squares or simple trinomials.
Example 3: Solve 2 sin x cos x − cos x = 0 on [0, 2π)
Step 1: Factor out the common factor cos x.
cos x (2 sin x − 1) = 0
Step 2: Set each factor equal to zero.
cos x = 0 or 2 sin x − 1 = 0
Step 3: Solve cos x = 0.
x = π/2 and x = 3π/2
Step 4: Solve 2 sin x − 1 = 0, which gives sin x = 1/2.
x = π/6 and x = 5π/6
Answer: x = π/6, π/2, 5π/6, 3π/2
Key Takeaway
Never divide both sides of a trig equation by a trig function (like dividing by cos x). You will lose solutions where that function equals zero. Always factor instead, then apply the zero-product property.
Using Identities to Solve Equations
When a trigonometric equation contains more than one type of trig function, you often need to use a trigonometric identity to rewrite the equation in terms of a single function. The most commonly used identities for solving equations are:
- Pythagorean identity: sin²x + cos²x = 1, which lets you replace sin²x with 1 − cos²x (or vice versa)
- Double-angle identities: sin 2x = 2 sin x cos x and cos 2x = 1 − 2 sin²x = 2 cos²x − 1
- Tangent-secant identity: 1 + tan²x = sec²x
Using the Pythagorean Identity
If your equation has both sin x and cos x terms (with at least one squared), substitute using sin²x = 1 − cos²x or cos²x = 1 − sin²x to convert the equation into a single variable. This usually produces a quadratic equation that you can factor or solve with the quadratic formula.
Using Double-Angle Identities
If an equation involves sin 2x or cos 2x alongside sin x or cos x, expand the double-angle expression to convert everything to the same angle. For example, replace cos 2x with 1 − 2 sin²x when you want everything in terms of sin x.
Example 4: Solve cos 2x + sin x = 0 on [0, 2π)
Step 1: Replace cos 2x with the identity 1 − 2 sin²x.
1 − 2 sin²x + sin x = 0
Step 2: Rearrange into standard quadratic form (multiply by −1 for convenience).
2 sin²x − sin x − 1 = 0
Step 3: Factor the quadratic.
(2 sin x + 1)(sin x − 1) = 0
Step 4: Solve each factor.
2 sin x + 1 = 0 → sin x = −1/2 → x = 7π/6, 11π/6
sin x − 1 = 0 → sin x = 1 → x = π/2
Answer: x = π/2, 7π/6, 11π/6
Quadratic-Form Trig Equations
A quadratic-form trig equation is an equation that looks like a quadratic polynomial, but with a trig function playing the role of the variable. These equations have the general form:
where "trig" stands for sin, cos, tan, or another trig function. To solve these, you can either factor directly or use the quadratic formula by making a temporary substitution like u = sin x.
Substitution Method
For complex quadratic-form equations, let u = sin x (or cos x, etc.), solve the quadratic in u, then translate back to find x. Remember to check that your solutions for u are within the valid range of the trig function (for sine and cosine, −1 ≤ u ≤ 1).
Example 5: Solve 2 sin²x + sin x − 1 = 0 on [0, 2π)
Step 1: Let u = sin x. The equation becomes 2u² + u − 1 = 0.
Step 2: Factor: (2u − 1)(u + 1) = 0.
Step 3: Solve for u.
2u − 1 = 0 → u = 1/2
u + 1 = 0 → u = −1
Step 4: Translate back to x.
sin x = 1/2 → x = π/6, 5π/6
sin x = −1 → x = 3π/2
Answer: x = π/6, 5π/6, 3π/2
Example 6: Solve 2 cos²x − 3 cos x + 1 = 0 on [0, 2π)
Step 1: Let u = cos x. The equation becomes 2u² − 3u + 1 = 0.
Step 2: Factor: (2u − 1)(u − 1) = 0.
Step 3: Solve for u.
2u − 1 = 0 → u = 1/2
u − 1 = 0 → u = 1
Step 4: Translate back to x.
cos x = 1/2 → x = π/3, 5π/3
cos x = 1 → x = 0
Answer: x = 0, π/3, 5π/3
Key Takeaway
When you see a trig function raised to the second power plus or minus a first-power trig term and a constant, treat it like a quadratic. Factor it or use the quadratic formula with a u-substitution. Always reject solutions where u falls outside the range of the trig function.
Multi-Step Strategies and Tips
Many exam-level trig equations require combining several techniques in sequence. Here is a systematic approach you can follow for any trigonometric equation:
- Simplify first. Combine like terms and move everything to one side so the equation equals zero.
- Reduce to one trig function. Use identities (Pythagorean, double-angle, etc.) to rewrite the equation in terms of a single trig function if possible.
- Factor or use the quadratic formula. Once you have a single trig function, look for factoring opportunities. If the expression does not factor neatly, apply the quadratic formula.
- Solve for the angle. Use inverse trig functions and reference angles to find all solutions in one period.
- Write the general solution or restrict to the given interval. Add the appropriate multiples of the period (2nπ for sine/cosine, nπ for tangent), or list solutions within the specified interval.
- Check for extraneous solutions. If you squared both sides at any point, or if domain restrictions apply, verify each solution in the original equation.
Common Mistakes to Avoid
- Dividing by a trig function: Never divide both sides by sin x, cos x, or any trig expression. You will lose solutions where that expression equals zero. Factor instead.
- Forgetting the range of sin and cos: If your quadratic gives sin x = 2 or cos x = −3, those have no real solutions. Discard them.
- Missing quadrants: Each basic trig value typically produces two solutions per period (except at boundary values like 0, ±1). Make sure you find both.
- Confusing equations with identities: An equation has specific solutions; an identity is always true. Do not try to "solve" an identity.
- Forgetting the period: When asked for all solutions, you must include the general term (+ 2nπ or + nπ). When restricted to an interval like [0, 2π), list only the values in that range.
When the Quadratic Formula is Needed
Not every quadratic trig equation factors nicely over the integers. When factoring fails, apply the quadratic formula directly. For example, given 3 sin²x − 5 sin x + 1 = 0, let u = sin x and compute:
Evaluate both values numerically and check whether each falls within [−1, 1]. If a value is valid, use arcsin to find the reference angle and determine all solutions in the required interval. If a value lies outside [−1, 1], discard it.
Practice Problems
Practice 1
Solve 2 cos²x − cos x = 0 on the interval [0, 2π).
Solution:
Factor out cos x:
cos x (2 cos x − 1) = 0
Set each factor to zero:
cos x = 0 → x = π/2, 3π/2
2 cos x − 1 = 0 → cos x = 1/2 → x = π/3, 5π/3
Answer: x = π/3, π/2, 3π/2, 5π/3
Practice 2
Find all solutions of tan²x − 3 = 0.
Solution:
Isolate tan²x:
tan²x = 3
tan x = ±√3
For tan x = √3: the reference angle is π/3. Since tangent has period π:
x = π/3 + nπ
For tan x = −√3: the reference angle is π/3 (in QII):
x = 2π/3 + nπ
General solution: x = π/3 + nπ or x = 2π/3 + nπ, where n is any integer.
(This can also be written as x = π/3 + nπ/1 combining both families: x = ±π/3 + nπ.)
Practice 3
Solve sin²x − sin x − 2 = 0 on [0, 2π).
Solution:
Let u = sin x. Then u² − u − 2 = 0.
Factor: (u − 2)(u + 1) = 0
u = 2 or u = −1
sin x = 2 has no solution (outside [−1, 1]).
sin x = −1 → x = 3π/2
Answer: x = 3π/2