Solving Trig Equations (AI HL)
Once a sine or cosine model is set up, exam questions usually ask you to run it backwards: given a height, a voltage, or a depth, find the time (or times) at which it occurs. Because trig functions repeat, a single equation can have several valid answers within the interval you're given. This page covers the graphical GDC method IB expects, plus the algebraic route for cases without a calculator. It's part of the broader Trigonometric Models topic.
21 questions on this sub-topic.
The method
Covered under IB syllabus reference AHL3.8: graphical methods of solving trigonometric equations in a finite interval. There's no single formula to memorise here - it's a technique, and the calculator does the heavy lifting.
Graphical (GDC) method
Graph \(y = f(x)\) and \(y = k\) (or the right-hand side of the equation) over the given interval, then use the intersect function to read off every \(x\)-value where the two graphs cross.
This is the standard IB approach for equations like \(a\sin(bx)+d=k\) once numbers get messy - fast, reliable, and it naturally shows you how many solutions to expect.
Algebraic method
Isolate the trig ratio, apply the inverse function for one solution, then use symmetry (\(\sin(180^\circ-\theta)=\sin\theta\) or the equivalent in radians) and add multiples of the period to find the rest within the interval.
Needed on non-calculator papers, or when a question asks for an exact answer rather than a decimal.
Want the sinusoidal model itself - amplitude, period, principal axis? See Trigonometric Models.
Worked examples
For \(h(t) = 3\sin\!\left(\tfrac{\pi}{6}t\right) + 5\).
(a)(i) Find the earlier time (\(t > 0\)) at which the height is 6.5 m.
(a)(ii) Find the later time.
Worked solution
(a)(i) \(3\sin\!\left(\tfrac{\pi}{6}t\right) = 1.5 \Rightarrow \sin\!\left(\tfrac{\pi}{6}t\right) = 0.5.\) M1 A1
\(\tfrac{\pi}{6}t = \tfrac{\pi}{6}\) or \(\tfrac{5\pi}{6}.\) M1 A1
\(t = 1\) A1
(a)(ii) or \(5\) h. A1
The depth of water, in metres, at a harbour is \(h(t) = 6 + 4\cos(30t^\circ)\), where \(t\) is hours after noon.
(a) State the maximum and minimum depth.
(b) Find the depth at \(t = 4\).
Worked solution
(a) Max \(= 6 + 4 = 10\) m; min \(= 6 - 4\) M1
\(= 2\) m. A1 A1
(b) \(h(4) = 6 + 4\cos(120^\circ)\) M1
\(= 4\) m. A1
Common mistakes
- Only finding one solution to a trig equation. Sine and cosine equations generally have multiple solutions within a given interval - always check the context (e.g. "the first two times") for how many values are needed.
- Forgetting to restrict the domain before solving. The calculator's intersect tool only finds crossings inside the graphing window - set the window to match the interval given in the question, or you'll miss (or invent) solutions.
- Mixing degree and radian mode. A model written with \(t^\circ\) needs the GDC in degree mode; one written as \(\tfrac{\pi}{6}t\) needs radian mode. Solving in the wrong mode gives numbers that look plausible but are wrong.
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Quick answers
How do I solve a trig equation on the GDC?
Graph the left-hand side as one function and the right-hand side as a second function (or a horizontal line), restrict the window to the given interval, and use the calculator's intersect tool to read off every crossing point.
Why do trig equations have more than one solution?
Sine and cosine repeat every period, so a horizontal line through the graph crosses it once for every full or partial cycle inside the given interval - always check the domain to see how many crossings to expect.