Grade by Grade
B. Trigonometric Functions · Trigonometric Identities and Equations

Solving Quadratic Trigonometric Equations by Factoring

B3.4About 75 minutes
Learning goals
  • I can recognize a trigonometric equation that is quadratic in sin x, cos x or tan x.
  • I can factor it (common factor, simple trinomial, difference of squares) or use the quadratic formula.
  • I can solve each factor for 0 ≤ x ≤ 2π and reject impossible values such as sin x = −2.
  • I can explain why I should factor instead of dividing by a trigonometric ratio.

Words to know

quadratic trigonometric equation
An equation that is quadratic in a trigonometric ratio, such as 2 sin²x − sin x − 1 = 0 (quadratic in sin x).
zero product property
If a · b = 0, then a = 0 or b = 0. It turns one factored equation into simpler equations.
extraneous or inadmissible value
A value from the algebra that cannot be used, such as sin x = −2 or cos x = 2.3, since sine and cosine lie between −1 and 1.
quadratic formula
For au² + bu + c = 0, u = (−b ± √(b² − 4ac))/(2a).

A quadratic in disguise

Look at 2 sin²x − sin x − 1 = 0. If you cover up "sin x" and write u instead, it becomes 2u² − u − 1 = 0, a quadratic you have factored many times since Grade 10. That is the whole idea of this lesson: treat the ratio as the variable, solve the quadratic, then solve each simpler trigonometric equation with the methods of lesson 10.

A quadratic can give two values of the ratio, and each value can give up to two angles, so expect up to four solutions in 0 ≤ x ≤ 2π (more if the angle is a multiple like 2x).

2 sin²x − sin x − 1 = 0Let u = sin x: 2u² − u − 1 = 0Factor: (2u + 1)(u − 1) = 0u = −1/2 or u = 1sin x = −1/2x = 7π/6, 11π/6sin x = 1x = π/2put sin x back
The substitution path: replace sin x by u, factor, solve for u, then put sin x back and solve each simple equation.
Worked example

Example 1: Factoring a trinomial

Solve 2 sin²x − sin x − 1 = 0 for 0 ≤ x ≤ 2π.

  1. Let u = sin x: 2u² − u − 1 = 0, which factors as (2u + 1)(u − 1) = 0.
  2. So 2 sin x + 1 = 0 or sin x − 1 = 0, giving sin x = −1/2 or sin x = 1.
  3. sin x = −1/2: α = π/6, quadrants III and IV: x = 7π/6, 11π/6.
  4. sin x = 1: x = π/2.

x = π/2, 7π/6, 11π/6

xπ/2π3π/22π21−1π/2 (touches)7π/611π/6y = 2sin²x − sin x − 1
The graph of y = 2sin²x − sin x − 1 meets the x-axis exactly at the three solutions. At π/2 it only touches, because the factor (sin x − 1) is never positive.
Worked example

Example 2: Difference of squares

Solve tan²x − 3 = 0 for 0 ≤ x ≤ 2π.

  1. (tan x − √3)(tan x + √3) = 0, so tan x = √3 or tan x = −√3.
  2. α = π/3 in both cases.
  3. tan x = √3 (quadrants I, III): π/3, 4π/3. tan x = −√3 (quadrants II, IV): 2π/3, 5π/3.

x = π/3, 2π/3, 4π/3, 5π/3

Worked example

Example 3: Common factor

Solve 2cos²x + cos x = 0 for 0 ≤ x ≤ 2π.

  1. Common factor: cos x(2 cos x + 1) = 0.
  2. cos x = 0: x = π/2, 3π/2.
  3. cos x = −1/2: α = π/3, quadrants II and III: x = 2π/3, 4π/3.

x = π/2, 2π/3, 4π/3, 3π/2

Worked example

Example 4: Never divide by a ratio

Solve 2 sin x cos x = sin x for 0 ≤ x ≤ 2π.

  1. Tempting but wrong: divide both sides by sin x to get 2 cos x = 1. This throws away every solution with sin x = 0.
  2. Correct: bring everything to one side and factor: 2 sin x cos x − sin x = 0, so sin x(2 cos x − 1) = 0.
  3. sin x = 0: x = 0, π, 2π.
  4. cos x = 1/2: x = π/3, 5π/3.

x = 0, π/3, π, 5π/3, 2π (dividing would have lost 0, π and 2π).

Worked example

Example 5: Rejecting an impossible value

Solve 3 sin²x + 5 sin x − 2 = 0 for 0 ≤ x ≤ 2π, to two decimal places.

  1. Factor: (3 sin x − 1)(sin x + 2) = 0. Check: 3 sin²x + 6 sin x − sin x − 2 ✓.
  2. sin x = 1/3: α = sin⁻¹(1/3) ≈ 0.3398, quadrants I and II: x ≈ 0.3398, π − 0.3398 ≈ 2.8018.
  3. sin x = −2: impossible, since −1 ≤ sin x ≤ 1. Reject.

x ≈ 0.34, 2.80

Worked example

Example 6: When it will not factor

Solve cos²x − 3 cos x − 1 = 0 for 0 ≤ x ≤ 2π, to two decimal places.

  1. Use the quadratic formula with u = cos x: u = (3 ± √(9 + 4))/2 = (3 ± √13)/2.
  2. (3 + √13)/2 ≈ 3.30: impossible for cosine. Reject.
  3. (3 − √13)/2 ≈ −0.3028. α = cos⁻¹(0.3028) ≈ 1.2632.
  4. Cosine is negative in quadrants II and III: x ≈ π − 1.2632 ≈ 1.8784 and x ≈ π + 1.2632 ≈ 4.4048.

x ≈ 1.88, 4.40

Checklist for quadratic trig equations

  • Move every term to one side, so the other side is 0.
  • Make sure there is only one ratio (sin x, or cos x, or tan x). If there are two, you may need an identity (next lesson).
  • Factor, or use the quadratic formula.
  • Reject values outside [−1, 1] for sine and cosine.
  • Solve each factor and list every solution once, in order.
Just for fun

Quadratic trig equations are ordinary quadratics wearing a disguise. Let u = sin x and the disguise falls off. Unlike a Halloween costume, though, you have to put it back on at the end.

Hands-on activity

Plot it and find the zeros

You need: graph paper, a pencil and a calculator in radian mode.

Step 1. Make a table of f(x) = 2sin²x − sin x − 1 for x = 0, 0.25, 0.5, ..., 6.25.

Step 2. Plot the points and join them smoothly.

Step 3. Read off where the curve meets the x-axis, and compare with Example 1: π/2 ≈ 1.57, 7π/6 ≈ 3.67 and 11π/6 ≈ 5.76.

What to look for: f(0) = −1; the curve rises to 0 at about 1.57, touches the axis there and turns back down; then crosses the axis at about 3.67 (going up) and 5.76 (going down). The highest values (2) occur at 3π/2 ≈ 4.71, where sin x = −1.

Your turn

1

Solve 2cos²x − cos x − 1 = 0 for 0 ≤ x ≤ 2π.

2

Solve 4sin²x − 3 = 0 for 0 ≤ x ≤ 2π.

3

Solve tan²x − tan x = 0 for 0 ≤ x ≤ 2π.

4

Solve 2sin²x + 3 sin x + 1 = 0 for 0 ≤ x ≤ 2π.

5

Solve cos²x = 1/2 for 0 ≤ x ≤ 2π.

6

Solve sin x tan x = sin x for 0 ≤ x ≤ 2π.

7

Solve 2tan²x + tan x − 1 = 0 for 0 ≤ x ≤ 2π (two decimal places where needed).

More practice: warm-up

1

Factor 2u² + u − 1.

2

Solve sin²x = 1 for 0 ≤ x ≤ 2π.

3

Solve cos²x = 1 for 0 ≤ x ≤ 2π.

4

Solve tan²x = 1 for 0 ≤ x ≤ 2π.

5

A factor gives cos x = 2. What do you do with it?

More practice: on your own

1

Solve 2sin²x = sin x for 0 ≤ x ≤ 2π.

2

Solve 6cos²x − cos x − 2 = 0 for 0 ≤ x ≤ 2π (two decimal places where needed).

3

Solve sec²x − 2 sec x = 0 for 0 ≤ x ≤ 2π.

4

Solve sin²x − 2 sin x − 1 = 0 for 0 ≤ x ≤ 2π, to two decimal places.

5

Solve 4cos³x − cos x = 0 for 0 ≤ x ≤ 2π.

Challenge

1

Solve sin²(2x) = 1/4 for 0 ≤ x ≤ 2π.

2

Solve 2 sin x cos x − √2 sin x + √2 cos x − 1 = 0 for 0 ≤ x ≤ 2π. (Hint: factor by grouping.)

3

The graph of y = 2sin²x − sin x − 1 touches the x-axis at π/2 but crosses it at 7π/6. Explain why, using the factored form.

Check your understanding

Quiz

1.

What is the best first step to solve 2sin²x − sin x − 1 = 0?

2.

What are the solutions of 2cos²x + cos x = 0 for 0 ≤ x ≤ 2π?

3.

Why should you not divide both sides of 2 sin x cos x = sin x by sin x?

4.

Factoring gives sin x = 1/2 or sin x = −2. Which statement is correct?

5.

How many solutions does tan²x = 3 have for 0 ≤ x ≤ 2π?

6.

The quadratic formula gives cos x ≈ 3.30 or cos x ≈ −0.30. What should you do?

For parents and teachers

Big idea (B3.4): a quadratic trigonometric equation is an ordinary quadratic in sin x, cos x or tan x. Factor (or use the quadratic formula), solve each factor with the methods from lesson 10, and reject impossible values like sin x = −2.

How to help: If factoring with sin x is confusing, have your teen literally write u = sin x, solve for u, and then write sin x = ... for each value. Ask them to predict the maximum number of solutions before solving.

Common difficulties:

  • Dividing both sides by sin x or cos x and losing solutions.
  • Keeping impossible values (sin x = 2).
  • Forgetting the ± when taking a square root (sin²x = 3/4 gives sin x = ±√3/2).
  • Listing a repeated solution twice, or missing endpoints 0 and 2π.