Periodic graphs cross a level many times: Because sin, cos and tan repeat, an equation like sin x = 0.5 has many solutions. A question fixes a domain (e.g. 0° ≤ x ≤ 360°) and wants all solutions in it — usually two for sin/cos.
Read the interval first: Underline the domain. The number of answers depends on it — over one full period sin = k (|k|<1) gives two solutions.
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First solution, then its partner: Take the inverse for the first angle, then use symmetry for the others: sin x = k → x and 180° − x; cos x = k → x and 360° − x; tan x = k → x and x + 180°.
IB-style question — two solutions
Solve sin x = 0.5 for 0° ≤ x ≤ 360°.
Step by step
- First solution.
- Second (sin: 180° − x).
Final answer
x = 30° or x = 150°.
Negative k changes the quadrants: For cos x = −0.5 the solutions are in Q2 and Q3; sketch the unit circle (or use CAST) to place them correctly.
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Stretch the interval, solve, then divide: For sin(2x) = k on 0° ≤ x ≤ 360°: let the inside be the new variable, solve over the stretched interval 0° ≤ 2x ≤ 720°, find all those solutions, then divide each by 2. A multiple angle gives more solutions.
IB-style question — sin(2x)
Solve sin(2x) = 0.5 for 0° ≤ x ≤ 360°.
Step by step
- Let u = 2x, so 0° ≤ u ≤ 720°. Solve sin u = 0.5.
- Divide each by 2.
Final answer
x = 15°, 75°, 195°, 255° (four solutions).
Don't divide too early: Find all the 2x solutions across the doubled interval first — if you halve before finding them all, you lose answers.
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Substitute, factor, then solve each: An equation like 2sin²x − sinx − 1 = 0 is a quadratic in sin x. Let s = sin x, factor/solve for s, then solve sin x = each value over the interval (rejecting any s outside [−1, 1]).
IB-style question — quadratic in sin
Solve 2sin²x − sinx − 1 = 0 for 0° ≤ x ≤ 360°.
Step by step
- Let s = sin x; factor.
- So sin x = 1 or sin x = −½.
Final answer
x = 90°, 210°, 330°.
Use a Pythagorean swap if mixed: If the equation mixes sin² and cos (e.g. 2cos²x + sinx = …), use cos²x = 1 − sin²x first to get a single ratio, then it's a quadratic.
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Graph and use intersect: On Paper 2, solve a trig equation by graphing each side and using intersect (or graphing the difference and finding zeros) over the given interval — ideal inside a sinusoidal model (e.g. 'when is the height 5 m?').
IB-style question — within a model
A tide height is h = 4 sin(30t)° + 6 (t in hours). Find the first time t > 0 when the height is 8 m.
Step by step
- Set up the equation.
- Graph y = h and y = 8 on the GDC and read the first intersection.
Final answer
First at t = 1 hour (the GDC intersect confirms it).
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Watch the interval and the mode: Set the GDC to the right angle mode and window the interval the question asks for — then read every intersection in range.