The Core Five-Step Strategy
Every trig equation in Stewart Chapter 7 — no matter how complex — reduces to the same five-step process. Master this sequence and every technique below is just a variation on how you reach Step 1.
Isolate
Algebraically manipulate until one trig function stands alone on one side.
Reference Angle
Apply arcsin, arccos, or arctan to the absolute value of the right side.
Quadrant Analysis
Determine which quadrants the sign of the value falls in.
List Solutions
Write all specific solutions in [0, 2pi) using the quadrant angles.
General Form
Add +2pi*n (or +pi*n for tan) for the complete infinite solution set.
Quadrant Sign Reference
| Function | Positive (value > 0) | Negative (value < 0) | Period | General solution adds |
|---|---|---|---|---|
| sin(theta) | Q1, Q2 (y-coord positive) | Q3, Q4 (y-coord negative) | 2pi | 2pi * n |
| cos(theta) | Q1, Q4 (x-coord positive) | Q2, Q3 (x-coord negative) | 2pi | 2pi * n |
| tan(theta) | Q1, Q3 | Q2, Q4 | pi | pi * n |
| csc(theta) | Q1, Q2 | Q3, Q4 | 2pi | 2pi * n |
| sec(theta) | Q1, Q4 | Q2, Q3 | 2pi | 2pi * n |
| cot(theta) | Q1, Q3 | Q2, Q4 | pi | pi * n |
Essential Unit Circle Values to Memorize
| Angle | sin | cos | tan |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| pi/6 (30 deg) | 1/2 | sqrt(3)/2 | 1/sqrt(3) |
| pi/4 (45 deg) | sqrt(2)/2 | sqrt(2)/2 | 1 |
| pi/3 (60 deg) | sqrt(3)/2 | 1/2 | sqrt(3) |
| pi/2 (90 deg) | 1 | 0 | undefined |
| 2pi/3 (120 deg) | sqrt(3)/2 | -1/2 | -sqrt(3) |
| 3pi/4 (135 deg) | sqrt(2)/2 | -sqrt(2)/2 | -1 |
| 5pi/6 (150 deg) | 1/2 | -sqrt(3)/2 | -1/sqrt(3) |
| pi (180 deg) | 0 | -1 | 0 |
| 7pi/6 (210 deg) | -1/2 | -sqrt(3)/2 | 1/sqrt(3) |
| 5pi/4 (225 deg) | -sqrt(2)/2 | -sqrt(2)/2 | 1 |
| 4pi/3 (240 deg) | -sqrt(3)/2 | -1/2 | sqrt(3) |
| 3pi/2 (270 deg) | -1 | 0 | undefined |
| 5pi/3 (300 deg) | -sqrt(3)/2 | 1/2 | -sqrt(3) |
| 7pi/4 (315 deg) | -sqrt(2)/2 | sqrt(2)/2 | -1 |
| 11pi/6 (330 deg) | -1/2 | sqrt(3)/2 | -1/sqrt(3) |
Linear Trigonometric Equations
Stewart Precalculus § 7.5
A linear trig equation has the trig function appearing to the first power only. The strategy is purely algebraic: isolate the trig function, identify the reference angle from the unit circle, determine the correct quadrants, and list every solution.
Example 1 — Basic Sine Equation
Solve: 2 sin(x) − 1 = 0 on [0, 2pi)
Isolate: 2 sin(x) = 1, so sin(x) = 1/2.
Reference angle: arcsin(1/2) = pi/6.
sin(x) = 1/2 > 0, so x is in Q1 or Q2. Q1 angle: pi/6. Q2 angle: pi − pi/6 = 5pi/6.
Solutions on [0, 2pi): x = pi/6 and x = 5pi/6.
General solution: x = pi/6 + 2pi*n or x = 5pi/6 + 2pi*n, where n is any integer.
Example 2 — Cosine, Negative Value
Solve: 2 cos(x) + sqrt(3) = 0 on [0, 2pi)
Isolate: cos(x) = −sqrt(3)/2.
Reference angle: arccos(sqrt(3)/2) = pi/6.
cos(x) < 0, so x is in Q2 or Q3. Q2 angle: pi − pi/6 = 5pi/6. Q3 angle: pi + pi/6 = 7pi/6.
Solutions on [0, 2pi): x = 5pi/6 and x = 7pi/6.
General solution: x = 5pi/6 + 2pi*n or x = 7pi/6 + 2pi*n.
Example 3 — Tangent Equation
Solve: tan(x) + 1 = 0 on [0, 2pi)
Isolate: tan(x) = −1.
Reference angle: arctan(1) = pi/4.
tan(x) < 0 in Q2 and Q4. Q2 angle: pi − pi/4 = 3pi/4. Q4 angle: 2pi − pi/4 = 7pi/4.
Solutions on [0, 2pi): x = 3pi/4 and x = 7pi/4.
Tangent has period pi, so the general solution compresses to one family: x = 3pi/4 + pi*n (which includes 7pi/4 when n = 1).
Key Insight: Why tan has one family but sin and cos have two
Sine and cosine have period 2pi and each value is achieved in exactly two quadrants per period (Q1+Q2 for positive sin, Q1+Q4 for positive cos, etc.). This gives two separate solution families, each shifted by 2pi*n. Tangent has period pi and each value is achieved in exactly one quadrant per period (Q1 or Q3), so all solutions collapse into one family shifted by pi*n.
Solutions on [0, 2pi) vs. General Solutions
Understanding what the problem is actually asking
Solutions on [0, 2pi)
This means restrict your answer to angles in exactly one full revolution: from 0 (included) up to but not including 2pi. You list a finite set of angles.
Problem language: “Find all solutions on [0, 2pi)” or “Solve for x in [0, 2pi)”.
General Solution
This gives ALL solutions across every rotation. You add integer multiples of the period to each base solution. The result is an infinite family of angles.
Problem language: “Find all solutions” or “Give the general solution”.
Worked Illustration: sin(x) = 1/2
On [0, 2pi):
Reference angle = pi/6. Q1 and Q2 give positive sin.
x = pi/6, x = 5pi/6
Just these two values.
General solution:
Same base angles, then add 2pi*n to each family.
x = pi/6 + 2pi*n
x = 5pi/6 + 2pi*n
Infinitely many values: pi/6, 5pi/6, 13pi/6, 17pi/6, -11pi/6, etc.
Common Mistake to Avoid
Never write “x = pi/6 + 2pi*n or x = 5pi/6 + 2pi*n” when the problem asks only for solutions on [0, 2pi). Conversely, if the problem asks for “all solutions,” never give only the two values in [0, 2pi) without the +2pi*n. Read the problem carefully every time.
Equations with Multiple Angles
Equations like cos(2x) = 1/2, sin(3x) = −sqrt(2)/2, tan(x/2) = 1
When the argument is a multiple of x, you solve for the full argument first, then divide by the coefficient. The key complication: dividing the period also changes how many base solutions exist in [0, 2pi). You must account for more rotations of the inner argument to capture all solutions.
The Multiple-Angle Rule
For sin(kx) or cos(kx), solve for u = kx first using u in [0, 2pi*k) — that is, extend the interval by a factor of k. Then divide by k. This ensures you capture all solutions in [0, 2pi) for x.
Equivalently: write the general solution for u first, then divide every term by k.
Example 4 — Double Angle
Solve: cos(2x) = 1/2 on [0, 2pi)
Since x is in [0, 2pi), u = 2x is in [0, 4pi). We need all solutions for cos(u) = 1/2 in [0, 4pi).
arccos(1/2) = pi/3. Cosine is positive in Q1 and Q4.
u = pi/3 and u = 5pi/3.
Add 2pi to each: u = pi/3 + 2pi = 7pi/3, u = 5pi/3 + 2pi = 11pi/3.
x = pi/6, x = 5pi/6, x = 7pi/6, x = 11pi/6.
Example 5 — Triple Angle
Solve: sin(3x) = −sqrt(2)/2 on [0, 2pi)
x in [0, 2pi) means u in [0, 6pi). Need all solutions in [0, 6pi).
arcsin(sqrt(2)/2) = pi/4. sin(u) < 0 means Q3 and Q4.
Q3: pi + pi/4 = 5pi/4. Q4: 2pi − pi/4 = 7pi/4. These repeat every 2pi, so in [0, 6pi) we get three full periods.
5pi/4, 7pi/4, 5pi/4+2pi=13pi/4, 7pi/4+2pi=15pi/4, 5pi/4+4pi=21pi/4, 7pi/4+4pi=23pi/4.
x = 5pi/12, 7pi/12, 13pi/12, 15pi/12=5pi/4, 21pi/12=7pi/4, 23pi/12.
Example 6 — Half Angle (x/2)
Solve: tan(x/2) = 1 on [0, 2pi)
x in [0, 2pi) means u in [0, pi). Only half a period.
tan(u) = 1. Reference angle = pi/4. tan is positive in Q1. u = pi/4.
Q3 gives u = pi/4 + pi = 5pi/4, but 5pi/4 is outside [0, pi), so it does not produce a solution for x in [0, 2pi).
u = pi/4 means x/2 = pi/4, so x = pi/2.
Factoring Trigonometric Equations
Quadratic substitution and direct factoring
When the equation is quadratic in a trig function, substitute u for the trig function, factor the resulting polynomial, and then solve each case. Always check that your u-solutions are in the valid range of the trig function.
Valid Ranges (Critical for Filtering)
Any u-solution outside these ranges produces no real angles and is discarded as extraneous.
Example 7 — Quadratic in Sine
Solve: sin²(x) + sin(x) − 2 = 0 on [0, 2pi)
Let u = sin(x). Equation becomes u² + u − 2 = 0.
(u + 2)(u − 1) = 0. So u = −2 or u = 1.
u = −2 is outside [−1, 1] for sine. Extraneous — discard. u = 1 is valid.
sin(x) = 1. On [0, 2pi): x = pi/2.
Example 8 — Quadratic in Cosine, Two Valid Solutions
Solve: 2cos²(x) − 3cos(x) + 1 = 0 on [0, 2pi)
Let u = cos(x). Equation: 2u² − 3u + 1 = 0.
(2u − 1)(u − 1) = 0. So u = 1/2 or u = 1.
Both 1/2 and 1 are in [−1, 1].
Reference angle pi/3. Q1 and Q4: x = pi/3 and x = 5pi/3.
x = 0 (the only angle in [0, 2pi) where cosine equals 1).
Example 9 — Direct Factoring (no substitution needed)
Solve: sin(x) * tan(x) − sin(x) = 0 on [0, 2pi)
sin(x) * [tan(x) − 1] = 0.
sin(x) = 0. On [0, 2pi): x = 0 and x = pi.
tan(x) = 1. Reference pi/4. tan positive in Q1 and Q3: x = pi/4 and x = 5pi/4.
Using Pythagorean Identities to Solve Equations
Converting between sin and cos to reach one function
sin²(x) + cos²(x) = 1
sin²(x) = 1 − cos²(x)
cos²(x) = 1 − sin²(x)
1 + tan²(x) = sec²(x)
tan²(x) = sec²(x) − 1
sec²(x) − tan²(x) = 1
1 + cot²(x) = csc²(x)
cot²(x) = csc²(x) − 1
csc²(x) − cot²(x) = 1
Example 10 — Mixed Sin and Cos, Use Pythagorean Identity
Solve: 2cos²(x) − sin(x) − 1 = 0 on [0, 2pi)
Replace cos²(x) with 1 − sin²(x):
(2sin(x) − 1)(sin(x) + 1) = 0.
sin(x) = 1/2. Q1 and Q2: x = pi/6 and x = 5pi/6.
sin(x) = −1. x = 3pi/2.
Example 11 — Tan and Sec via Pythagorean Identity
Solve: tan²(x) − sec(x) − 1 = 0 on [0, 2pi)
tan²(x) = sec²(x) − 1. Replace:
(sec(x) − 2)(sec(x) + 1) = 0.
sec(x) = 2 means cos(x) = 1/2. Q1 and Q4: x = pi/3 and x = 5pi/3.
sec(x) = −1 means cos(x) = −1. x = pi.
Squaring Both Sides — and Checking for Extraneous Solutions
An essential technique that always requires verification
Mandatory Verification Step
Squaring both sides of an equation is equivalent to multiplying both sides by the same quantity — but that quantity can be negative, which reverses the equality direction. As a result, the squared equation may have solutions that the original equation does not. You MUST substitute every candidate solution back into the original (unsquared) equation and reject any that do not satisfy it.
Example 12 — Squaring to Eliminate a Sum
Solve: sin(x) + cos(x) = 1 on [0, 2pi)
sin²(x) + 2sin(x)cos(x) + cos²(x) = 1. Since sin² + cos² = 1: 1 + 2sin(x)cos(x) = 1.
2sin(x)cos(x) = 0. Using the double-angle identity: sin(2x) = 0.
2x = 0, pi, 2pi, 3pi. So x = 0, pi/2, pi, 3pi/2.
x = 0: sin(0) + cos(0) = 0 + 1 = 1. Valid.
x = pi/2: sin(pi/2) + cos(pi/2) = 1 + 0 = 1. Valid.
x = pi: sin(pi) + cos(pi) = 0 + (−1) = −1. Not 1. Extraneous — reject.
x = 3pi/2: sin(3pi/2) + cos(3pi/2) = −1 + 0 = −1. Not 1. Extraneous — reject.
Sum-to-Product and Product-to-Sum Equations
Stewart Precalculus § 7.4
These identities convert sums of trig functions into products (and vice versa), which can make equations easier to factor and solve. They appear in Stewart Chapter 7 and on standardized exams.
Sum-to-Product
sin A + sin B = 2 sin((A+B)/2) cos((A-B)/2)
sin A − sin B = 2 cos((A+B)/2) sin((A-B)/2)
cos A + cos B = 2 cos((A+B)/2) cos((A-B)/2)
cos A − cos B = −2 sin((A+B)/2) sin((A-B)/2)
Product-to-Sum
sin A cos B = (1/2)[sin(A+B) + sin(A-B)]
cos A cos B = (1/2)[cos(A-B) + cos(A+B)]
sin A sin B = (1/2)[cos(A-B) − cos(A+B)]
Example 13 — Sum-to-Product to Factor
Solve: sin(3x) + sin(x) = 0 on [0, 2pi)
sin(3x) + sin(x) = 2 sin((3x+x)/2) cos((3x−x)/2) = 2 sin(2x) cos(x).
2 sin(2x) cos(x) = 0, so sin(2x) = 0 or cos(x) = 0.
2x = 0, pi, 2pi, 3pi. x = 0, pi/2, pi, 3pi/2.
x = pi/2 and x = 3pi/2. (Already included above.)
Graphical Interpretation of Trig Equations
Understanding solutions as intersections
Every trig equation of the form f(x) = c can be visualized as the intersection of the curve y = f(x) and the horizontal line y = c. The x-coordinates of the intersections in [0, 2pi) are your solutions. This graphical view builds intuition for why there are typically two solutions per period for sin and cos, and one for tan.
sin(x) = c on [0, 2pi)
- c = 1: one intersection at x = pi/2
- 0 < c < 1: two intersections in Q1 and Q2
- c = 0: two intersections at x = 0 and x = pi
- −1 < c < 0: two intersections in Q3 and Q4
- c = −1: one intersection at x = 3pi/2
- |c| > 1: no intersections (no solutions)
cos(x) = c on [0, 2pi)
- c = 1: one intersection at x = 0
- 0 < c < 1: two intersections in Q1 and Q4
- c = 0: two intersections at x = pi/2 and x = 3pi/2
- −1 < c < 0: two intersections in Q2 and Q3
- c = −1: one intersection at x = pi
- |c| > 1: no intersections (no solutions)
Using Graphs to Check Your Algebraic Work
After solving algebraically, use a graphing calculator to confirm:
- Graph y = f(x) and y = c in the window [0, 2pi) on the x-axis.
- Count the intersection points — this should match the number of solutions you found.
- Use the INTERSECT function to find x-coordinates and verify they match your algebraic answers.
- If the counts disagree, you either missed a quadrant or made an algebraic error.
Calculator-Based Solutions: Arcsin, Arccos, Arctan and Adjustments
For equations that do not have exact unit-circle answers
Most real-world and exam problems involving equations like sin(x) = 0.7 or cos(x) = −0.3 do not have exact radian answers. You use the inverse trig keys on a calculator to get one value, then apply quadrant analysis to find all others.
The Calculator Protocol
Get the reference value
Press arcsin, arccos, or arctan of the absolute value of the right side. The calculator returns a value in the principal range: arcsin in [-pi/2, pi/2], arccos in [0, pi], arctan in (-pi/2, pi/2).
Determine correct quadrants
Use the sign of the right side and the ASTC rule (All Students Take Calculus: All positive in Q1, Sin in Q2, Tan in Q3, Cos in Q4).
Compute all solutions in [0, 2pi)
Q1: use reference value as-is. Q2: pi minus reference. Q3: pi plus reference. Q4: 2pi minus reference. Select only the quadrants from Step 2.
Add period for general solution
Append + 2pi*n to each solution family. For tan equations, append + pi*n.
Example 14 — Non-Exact Sine Equation
Solve: sin(x) = 0.7 on [0, 2pi) to 4 decimal places
arcsin(0.7) ≈ 0.7754 radians.
sin(x) > 0 means Q1 and Q2.
x ≈ 0.7754
x ≈ pi − 0.7754 ≈ 3.1416 − 0.7754 = 2.3662
Example 15 — Non-Exact Cosine, Negative Value
Solve: cos(x) = −0.45 on [0, 2pi) to 4 decimal places
arccos(0.45) ≈ 1.1040 radians.
cos(x) < 0 means Q2 and Q3.
x ≈ pi − 1.1040 ≈ 2.0376
x ≈ pi + 1.1040 ≈ 4.2456
Common Mistakes and How to Avoid Them
Forgetting to add + 2pi*n (or + pi*n for tan)
Why it happens: Trig functions are periodic. Every base solution repeats infinitely. The general solution requires the period term.
Fix: Always write the general solution with n as an integer, even if the problem asks for solutions on [0, 2pi). Then restrict n to get only the interval solutions.
Finding only one solution when there should be two
Why it happens: Arcsin and arccos each return one value. But every equation sin(x) = c with |c| < 1 has two solutions per period.
Fix: After getting the reference angle, always ask: in which two quadrants does the sign match? List both solutions.
Dividing both sides by a trig function
Why it happens: If the trig function equals zero at some angle, dividing by it removes those solutions entirely.
Fix: Move everything to one side and factor. Never divide by sin(x), cos(x), or tan(x).
Forgetting to divide the period when using multiple angles
Why it happens: For cos(2x) = c, the argument is 2x, so the period of 2x is 2pi. But the period of x in the original equation is pi. Failing to extend the search interval misses solutions.
Fix: For k*x in the argument, solve for u = k*x in [0, 2pi*k), then divide by k.
Not checking extraneous solutions after squaring
Why it happens: Squaring can introduce solutions of the squared equation that do not satisfy the original.
Fix: Substitute every candidate solution into the original equation. Reject any that fail.
Treating arcsin as the inverse of the equation rather than finding a reference angle
Why it happens: arcsin only returns values in [-pi/2, pi/2]. For Q2 solutions of sin equations, you must compute pi minus arcsin(c), not arcsin(c) directly.
Fix: Always use arcsin, arccos, arctan only to get the reference angle. Then apply quadrant rules separately.
Using degree mode instead of radian mode on the calculator
Why it happens: Most precalculus and calculus work uses radians. An answer in degrees will be wrong.
Fix: Always verify the calculator is in radian mode before beginning any trig computation.
Quick-Reference: Strategy by Equation Type
| Equation Type | Example | Strategy |
|---|---|---|
| Linear in one trig function | 3sin(x) - 1 = 0 | Isolate, reference angle, quadrant analysis |
| Multiple angle | cos(2x) = 1/2 | Solve for the argument in an extended interval, then divide |
| Quadratic in one trig function | 2sin²(x) - sin(x) - 1 = 0 | Substitute u, factor polynomial, filter by range, solve each |
| Mixed sin and cos (no squares) | sin(x) = cos(x) | Divide both sides by cos(x) to get tan(x) = 1 (check cos ≠ 0) |
| Mixed sin and cos (squared) | 2cos²(x) - sin(x) - 1 = 0 | Replace cos² = 1 - sin², collect, factor |
| Sum of two trig functions = 0 | sin(3x) + sin(x) = 0 | Apply sum-to-product identity, factor, solve each factor |
| Equation with squaring needed | sin(x) + cos(x) = 1 | Square both sides, use Pythagorean identity, verify all solutions |
| Non-exact values | sin(x) = 0.7 | Calculator: arcsin for reference angle, adjust for both quadrants |
| Involves sec or csc | sec(x) - 2 = 0 | Rewrite in terms of cos or sin, then proceed normally |
| Product of trig functions = 0 | sin(x)tan(x) - sin(x) = 0 | Factor out common trig expression, set each factor to zero |
Additional Worked Examples
Example 16 — Mixed Equation: Divide to Get Tangent
Solve: sin(x) = cos(x) on [0, 2pi)
Divide both sides by cos(x): tan(x) = 1. (Note: cos(x) ≠ 0 at the solutions we find, so this division is valid here. Always verify afterwards.)
tan(x) = 1. Reference pi/4. Positive in Q1 and Q3.
x = pi/4 and x = 5pi/4.
At x = pi/4: sin(pi/4) = cos(pi/4) = sqrt(2)/2. Valid. At x = 5pi/4: sin(5pi/4) = cos(5pi/4) = −sqrt(2)/2. Valid.
Example 17 — Double Angle Identity Substitution
Solve: sin(2x) = cos(x) on [0, 2pi)
Use sin(2x) = 2sin(x)cos(x). Equation becomes 2sin(x)cos(x) = cos(x).
2sin(x)cos(x) − cos(x) = 0. Factor: cos(x)[2sin(x) − 1] = 0.
cos(x) = 0. x = pi/2 and x = 3pi/2.
2sin(x) − 1 = 0, sin(x) = 1/2. x = pi/6 and x = 5pi/6.
Example 18 — Equation Requiring Pythagorean Identity with Cosine Double Angle
Solve: cos(2x) + sin(x) = 0 on [0, 2pi)
Use cos(2x) = 1 − 2sin²(x). Substituting: 1 − 2sin²(x) + sin(x) = 0.
2sin²(x) − sin(x) − 1 = 0.
(2sin(x) + 1)(sin(x) − 1) = 0.
sin(x) = −1/2. Q3: 7pi/6. Q4: 11pi/6.
sin(x) = 1. x = pi/2.
Example 19 — Equation with Cosine Difference of Angles
Solve: cos(x − pi/3) = sqrt(3)/2 on [0, 2pi)
Let u = x − pi/3. Solve cos(u) = sqrt(3)/2.
arccos(sqrt(3)/2) = pi/6. Cosine positive in Q1 and Q4.
u = pi/6 + 2pi*n or u = −pi/6 + 2pi*n (equivalently 11pi/6 + 2pi*n).
x = u + pi/3. Family 1: x = pi/6 + pi/3 = pi/2. Family 2: x = −pi/6 + pi/3 = pi/6. Also check n=1 for family 2: x = −pi/6 + pi/3 + 2pi = pi/6 + 2pi (out of range). For family 1, n=1: x = pi/2 + 2pi (out of range).
Example 20 — Complex Quadratic with Cos Double Angle (Alternative Form)
Solve: cos(2x) − cos(x) = 0 on [0, 2pi)
Use cos(2x) = 2cos²(x) − 1: 2cos²(x) − 1 − cos(x) = 0. So 2cos²(x) − cos(x) − 1 = 0.
(2cos(x) + 1)(cos(x) − 1) = 0.
cos(x) = −1/2. Reference pi/3. Q2: 2pi/3. Q3: 4pi/3.
cos(x) = 1. x = 0.
Frequently Asked Questions
Q: What is the difference between solutions on [0, 2pi) and the general solution?
- ‣Solutions on [0, 2pi) are a finite set of specific angles in exactly one full rotation that satisfy the equation.
- ‣The general solution adds integer multiples of the period (2pi*n for sin and cos, pi*n for tan) to capture all infinitely many angles that work.
- ‣Read the problem carefully: 'find all solutions' means you need the general form. 'Find solutions on [0, 2pi)' means just the finite interval list.
Q: How do you solve a trig equation with a multiple angle like cos(2x) = 1/2?
- ‣Let u = 2x. Since x is in [0, 2pi), u is in [0, 4pi) — two full revolutions. Solve cos(u) = 1/2 in that extended interval: u = pi/3, 5pi/3, 7pi/3, 11pi/3.
- ‣Divide every solution by 2: x = pi/6, 5pi/6, 7pi/6, 11pi/6.
- ‣For the general solution, divide the period term as well: x = pi/6 + pi*n or x = 5pi/6 + pi*n.
Q: How do you factor a quadratic trig equation like sin squared x plus sin x minus 2 equals 0?
- ‣Substitute u = sin(x) to get the polynomial u^2 + u - 2 = 0.
- ‣Factor: (u + 2)(u - 1) = 0, giving u = -2 or u = 1.
- ‣Since sin(x) only takes values in [-1, 1], discard u = -2 as extraneous.
- ‣Solve sin(x) = 1: x = pi/2. General solution: x = pi/2 + 2pi*n.
Q: When do you need to check for extraneous solutions in trig equations?
- ‣Always check after squaring both sides. Squaring can create false solutions.
- ‣Also check when dividing — if you divided by a trig expression, verify the solutions do not make that expression zero.
- ‣The check is simple: substitute each candidate solution into the original (unsquared, undivided) equation and confirm equality.
Q: How do you use a calculator to solve trig equations that do not have exact answers?
- ‣Use arcsin, arccos, or arctan to get a reference value from the absolute value of the right side.
- ‣Determine which quadrants apply based on the sign. Compute solutions in those quadrants using pi minus, pi plus, or 2pi minus the reference value as appropriate.
- ‣Always verify the calculator is in radian mode. Append + 2pi*n for the general solution.
Q: What are the most common mistakes when solving trig equations?
- ‣Forgetting + 2pi*n or + pi*n for the general solution.
- ‣Missing the second quadrant solution (only writing one angle from arcsin or arccos).
- ‣Dividing both sides by a trig function, which loses the zero solutions of that function.
- ‣Not extending the interval when solving for multiple angles like 2x or 3x.
- ‣Forgetting to check for extraneous solutions after squaring.
Q: How do you use Pythagorean identities to solve trig equations?
- ‣When the equation mixes sin and cos or contains squared trig functions, use sin^2(x) + cos^2(x) = 1 to rewrite everything in terms of one function.
- ‣For example: 2cos^2(x) - sin(x) - 1 = 0. Replace cos^2(x) with 1 - sin^2(x) to get a quadratic in sin(x). Factor and solve.
- ‣The variant identities 1 + tan^2(x) = sec^2(x) and 1 + cot^2(x) = csc^2(x) work the same way for equations involving those functions.
Put This Into Practice
NailTheTest generates unlimited trig equation problems with step-by-step solutions and instant feedback. Work through every equation type in this guide until it feels automatic.
Practice Trig Equations Free