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Multiple Solutions and General Solutions

A free Trigonometry lesson from the “Solving Trig Equations” unit, with a worked example and practice problems including step-by-step solutions.

General solutions describe every repeated answer, often with +2pi k or +pi k for any integer k. In this lesson, the goal is to write repeated solutions using integer parameters. Prerequisite check: Algebra II or College Algebra foundations. Example 1: sin(x) = 1/2 on 0 <= x <= 2pi has solutions pi/6 and 5pi/6. Example 2: tan(x) = 1 repeats every pi, so a general solution is x = pi/4 + pi k. A common mistake is stopping after the first angle that works; the safer habit is to solve the basic trig value on the unit circle, then use symmetry and period.

What you'll learn

Why it matters: Trig equations appear whenever a repeated cycle reaches the same height, position, voltage, or angle more than once.

Worked example

Problem. Example 1 Foundation: What should you check before finalizing an interval-based trig equation?

  1. Intervals limit which repeated answers count.
  2. After finding candidate solutions, keep only those in the interval.
  3. This prevents extra or missing answers.

Answer: that every listed solution lies in the interval

Practice problems

1. Practice 1 Foundation: On 0 <= x <= 2pi, solve sin(x) = 1/2.

Choices: x = pi/6 and 5pi/6 · x = pi/6 only · x = 7pi/6 and 11pi/6 · x = pi/3 and 2pi/3

Show solution
  1. Warm-up: First identify exactly what the question is asking: Practice 1 Foundation: On 0 <= x <= 2pi, solve sin(x) = 1/2.
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. Sine is positive in Quadrants I and II.
  4. The reference angle is pi/6.
  5. So the interval solutions are pi/6 and 5pi/6.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: x = pi/6 and 5pi/6

2. Practice 2 Setup: On 0 <= x <= 2pi, solve cos(x) = 0.

Choices: x = pi/2 and 3pi/2 · x = 0 and pi · x = pi only · x = pi/4 and 5pi/4

Show solution
  1. Warm-up: First identify exactly what the question is asking: Practice 2 Setup: On 0 <= x <= 2pi, solve cos(x) = 0.
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. Cosine is the x-coordinate.
  4. x-coordinate 0 occurs on the y-axis.
  5. That gives pi/2 and 3pi/2.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: x = pi/2 and 3pi/2

3. Practice 3 Meaning: A general solution to tan(x) = 1 is:

Choices: x = pi/4 + pi k · x = pi/4 + 2pi k · x = pi/2 + pi k · x = pi k

Show solution
  1. Core Practice: First identify exactly what the question is asking: Practice 3 Meaning: A general solution to tan(x) = 1 is:
  2. For data questions, identify what each statistic measures before calculating so the result matches the question.
  3. tan(x) = 1 at pi/4.
  4. Tangent repeats every pi.
  5. So add pi k for any integer k.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: x = pi/4 + pi k

4. Practice 4 Method: Why can sin(x) = 1/2 have two answers on 0 <= x <= 2pi?

Choices: the same y-coordinate occurs in two quadrants · sine has no period · the reference angle changes size · there is only one unit-circle point

Show solution
  1. Core Practice: First identify exactly what the question is asking: Practice 4 Method: Why can sin(x) = 1/2 have two answers on 0 <= x <= 2pi?
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. Sine is the y-coordinate.
  4. A positive y-value appears in Quadrants I and II.
  5. Both angles can satisfy the equation.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: the same y-coordinate occurs in two quadrants

5. Practice 5 Reasoning: Solve sin^2(x) = 1 on 0 <= x <= 2pi.

Choices: x = pi/2 and 3pi/2 · x = 0 and pi · x = pi/4 and 5pi/4 · x = 2pi only

Show solution
  1. Core Practice: First identify exactly what the question is asking: Practice 5 Reasoning: Solve sin^2(x) = 1 on 0 <= x <= 2pi.
  2. Use inverse operations to isolate the unknown, and keep both sides balanced at every step.
  3. sin^2(x) = 1 means sin(x) = 1 or -1.
  4. Those occur at the top and bottom of the unit circle.
  5. So x = pi/2 and 3pi/2.
  6. Verify the selected choice by checking that it satisfies the original prompt and that the other choices fail the same test.

Answer: x = pi/2 and 3pi/2

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