A trigonometric graph lets you see where a function reaches a chosen level. The challenge is finding every permitted crossing, including those in a later cycle. A single calculator inverse value is a starting clue, not automatically the whole answer.
Every angle and horizontal coordinate in this lesson is in degrees. Use degree mode, label the horizontal axis with degree symbols and copy the interval exactly. Closed endpoints, written with ≤, are allowed; endpoints next to < are excluded.
Plan the graph from its amplitude and period
For y = A sin(Bx) or y = A cos(Bx), with A and B nonzero, the amplitude is |A| and the period is 360°/|B|. Amplitude is measured vertically from the midline, which is y = 0 in these examples. Period is the horizontal distance through one complete repeating cycle.
Use quarter-cycle landmarks to build a sketch. Sine progresses through 0, 1, 0, −1, 0 over a cycle; cosine progresses through 1, 0, −1, 0, 1. Multiply the heights by A and place them at horizontal spacings of one quarter of the period. Draw a smooth curve, not a chain of straight segments.
Case 1: compression and all level intersections
Solve 2 sin(2x) = 1 for 0° ≤ x ≤ 360°. The graph has amplitude 2 and period 360°/2 = 180°. The interval therefore contains two complete cycles, so stopping after the first two crossings would miss solutions.
| x | 0° | 45° | 90° | 135° | 180° | 225° | 270° | 315° | 360° |
|---|---|---|---|---|---|---|---|---|---|
| 2x | 0° | 90° | 180° | 270° | 360° | 450° | 540° | 630° | 720° |
| y = 2 sin(2x) | 0 | 2 | 0 | −2 | 0 | 2 | 0 | −2 | 0 |
The graph suggests x = 15°, 75°, 195° and 255°. Verify them without measuring the drawing. Divide by 2 to obtain sin(2x) = 1/2. Put θ = 2x, so the interval becomes 0° ≤ θ ≤ 720°.
In the first cycle, sine is 1/2 at θ = 30° and 150°. Adding 360° gives θ = 390° and 510° in the second cycle. Dividing all four by 2 gives the listed x-values. Substitution produces 2 sin 30°, 2 sin 150°, 2 sin 390° and 2 sin 510°, all equal to 1.
Inspect the endpoints separately: x = 0° gives 0 and x = 360° gives 2 sin 720° = 0. Neither equals 1. The solution set therefore has exactly four entries.
Case 2: a stretched cosine graph
Solve 3 cos(x/2) = −1.5 on 0° ≤ x ≤ 720°. Here the amplitude is 3 and the period is 360°/(1/2) = 720°. The horizontal division by 2 stretches the cycle across twice the usual degree interval.
| x | 0° | 180° | 360° | 540° | 720° |
|---|---|---|---|---|---|
| x/2 | 0° | 90° | 180° | 270° | 360° |
| y = 3 cos(x/2) | 3 | 0 | −3 | 0 | 3 |
Algebraically, cos(x/2) = −1/2. Set θ = x/2, giving 0° ≤ θ ≤ 360°. Cosine is negative in the second and third quadrants, and the reference angle is 60°, so θ = 120° or 240°. Multiply by 2: x = 240° or 480°.
Check 3 cos 120° = −1.5 and 3 cos 240° = −1.5. Neither endpoint works because cos 0° and cos 360° are both 1. This time two solutions are complete because the transformed-angle interval contains only one full cycle.
Case 3: keep the phase-shift brackets
Sketch y = sin[2(x − 30°)] and solve y = 0 for 0° ≤ x ≤ 360°. The amplitude is 1, the period is 180° and the graph is shifted 30° to the right compared with sin(2x). The bracket shows what happens to x before the factor 2 acts.
| x | 0° | 30° | 75° | 120° | 165° | 210° | 255° | 300° | 345° | 360° |
|---|---|---|---|---|---|---|---|---|---|---|
| 2(x − 30°) | −60° | 0° | 90° | 180° | 270° | 360° | 450° | 540° | 630° | 660° |
| y | −√3/2 | 0 | 1 | 0 | −1 | 0 | 1 | 0 | −1 | −√3/2 |
Put θ = 2(x − 30°). The given interval becomes −60° ≤ θ ≤ 660°. Sine is zero at integer multiples of 180°. The multiples in this transformed interval are 0°, 180°, 360° and 540°. Solve x = θ/2 + 30° to obtain 30°, 120°, 210° and 300°.
The next multiple, 720°, would give x = 390°, outside the interval. The previous multiple, −180°, would give x = −60°, also outside. At both original endpoints the function is −√3/2, not zero. These checks establish that no boundary or extra-cycle answer is missing.
Diagnose two common errors
Missing a cycle: for Case 1, writing only 15° and 75° means solving θ only up to 360° even though θ = 2x reaches 720°. Transform the whole interval when introducing a new angle, or add the x-period 180° to each first-cycle solution and test the result.
Wrong calculator mode: check sin 30° before a degree-based calculation; it should be 0.5. If the mode is radians, interpreting the number 30 produces a different angle. Also enter the entire bracketed argument in Case 3. The expressions sin(2x − 60°) and sin[2(x − 30°)] agree, but sin(2x) − 60 does not describe the same function.
Three solved practice tasks
- Solve 3 sin x = 1.5 for 0° ≤ x ≤ 540°. Sin x = 1/2 gives 30° and 150° in the first cycle. Add 360° to obtain 390° and 510°, both allowed. No endpoint works. The complete set is 30°, 150°, 390°, 510°.
- Solve −2 cos(3x) = 1 for 0° ≤ x ≤ 240°. Cos(3x) = −1/2 and 0° ≤ 3x ≤ 720°. Thus 3x = 120°, 240°, 480°, 600°, giving x = 40°, 80°, 160°, 200°. The negative coefficient changes vertical orientation, while amplitude remains 2.
- Solve sin x = 0 on 0° ≤ x ≤ 360°, then on 0° < x < 360°. The closed-interval answers are 0°, 180° and 360°. Removing both endpoints leaves only 180°. Repeated function values do not make different x-values duplicates.
Record the amplitude, period, transformed interval and endpoint checks beside your solution list. A labelled sketch explains how many crossings to expect; substitution confirms that each proposed crossing actually solves the equation.
Frequently asked questions
Why must I state degree mode?
The formulas and intervals in this lesson use degrees. A calculator interpreting 30 as radians evaluates a different angle, so even a correctly entered expression gives the wrong value for this task.
How do I find the period of A sin(Bx) or A cos(Bx)?
For nonzero A and B, with x in degrees, the period is 360°/|B|. A larger |B| gives a shorter period; a smaller nonzero |B| gives a longer period.
Can amplitude be negative?
No. Amplitude is a distance from the midline, so it is |A|. A negative coefficient reverses the graph vertically without making that distance negative.
Why is inverse sine not enough to solve an interval equation?
It supplies a principal angle, not all angles with the same sine. You must include the other appropriate quadrant, later or earlier cycles and the stated interval boundaries.
Does 2(x − 30°) shift a sine graph by 60°?
No. The bracketed form shows a shift of x by 30° to the right. Expanding gives 2x − 60°, but the horizontal shift is 60° divided by 2, which is 30°.
Should I count both ends of a closed interval?
Check them separately. For sin x = 0 on 0° ≤ x ≤ 360°, both 0° and 360° are distinct permitted x-values even though they produce the same function value.
What if the requested horizontal level lies above the maximum?
There are no intersections and no solutions in the interval. For example, 2 sin(2x) cannot equal 3 because its values lie between −2 and 2.
Further reading
For the underlying mathematics, see OpenStax’s sine and cosine graph treatment, expressed here in degrees. The mathematics topic-test collection is a separate resource page that currently shows a member login; specific files and their answers are not verified here.
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