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Solving on a Given Interval

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

Solving on a given interval means finding ALL angles in a stated window such as [0,2pi) or [0,360) that make a basic trig equation true. Because sine, cosine, and tangent repeat their values, a single equation usually has two solutions per cycle: you find the reference angle from the size of the value, then place a solution in each quadrant where the function has the required sign. The job is to list every solution in the interval and report how many there are.

What you'll learn

Why it matters: When a Ferris wheel, tide, or AC voltage repeats, asking "at what times in one full cycle is the height (or voltage) exactly this value?" is exactly an interval-solution problem, and the answer is usually two moments per turn, not one.

Worked example

Problem. Solve sin(x) = sqrt(3)/2 on 0 <= x < 2pi. List all solutions.

  1. The reference angle for sqrt(3)/2 is pi/3.
  2. Sine is positive in Quadrants I and II, so use pi/3 and pi - pi/3.
  3. The two solutions in the interval are pi/3 and 2pi/3.

Answer: pi/3, 2pi/3

Practice problems

1. Solve sin(x) = 1/2 on 0 <= x < 2pi. List all solutions.

Show solution
  1. Warm-up: First identify exactly what the question is asking: Solve sin(x) = 1/2 on 0 <= x < 2pi. List all solutions.
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. The reference angle for 1/2 is pi/6.
  4. Sine is positive in Quadrants I and II.
  5. The solutions are pi/6 and 5pi/6.
  6. Check the result by substituting or estimating: the response should match pi/6, 5pi/6 and make sense in the original problem.

Answer: pi/6, 5pi/6

2. Solve cos(x) = 1/2 on 0 <= x < 2pi. List all solutions.

Show solution
  1. Warm-up: First identify exactly what the question is asking: Solve cos(x) = 1/2 on 0 <= x < 2pi. List all solutions.
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. The reference angle for 1/2 is pi/3.
  4. Cosine is positive in Quadrants I and IV.
  5. The solutions are pi/3 and 5pi/3.
  6. Check the result by substituting or estimating: the response should match pi/3, 5pi/3 and make sense in the original problem.

Answer: pi/3, 5pi/3

3. How many solutions does sin(x) = 1 have on 0 <= x < 2pi?

Show solution
  1. Sine reaches its maximum value 1 only at the top of the unit circle.
  2. That happens once, at x = pi/2.
  3. So there is exactly 1 solution.

Answer: 1

4. Solve sin(x) = -1/2 on 0 <= x < 2pi. List all solutions.

Show solution
  1. Core Practice: First identify exactly what the question is asking: Solve sin(x) = -1/2 on 0 <= x < 2pi. List all solutions.
  2. For fractions, use equivalent forms, common denominators, or reciprocals depending on the operation being used.
  3. The reference angle for 1/2 is pi/6.
  4. Sine is negative in Quadrants III and IV.
  5. The solutions are 7pi/6 and 11pi/6.
  6. Check the result by substituting or estimating: the response should match 7pi/6, 11pi/6 and make sense in the original problem.

Answer: 7pi/6, 11pi/6

5. Solve cos(x) = -sqrt(2)/2 on 0 <= x < 2pi. List all solutions.

Show solution
  1. The reference angle for sqrt(2)/2 is pi/4.
  2. Cosine is negative in Quadrants II and III.
  3. The solutions are 3pi/4 and 5pi/4.

Answer: 3pi/4, 5pi/4

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