Solving Trigonometric Equations Graphically (within a Domain)
Solving trigonometric equations graphically uses the graphs of \(y=\sin x\), \(y=\cos x\) and \(y=\tan x\) to find every solution of an equation within a given domain by reading where the curves meet.
Part of the NSW Year 12 Mathematics Advanced course, in the Functions area of study (Further graph transformations and modelling focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.
Theory
Solving a trigonometric equation within a domain means finding every value of \(x\) in the interval that satisfies it. Graphically the solutions are where the curve \(y=f(x)\) meets the line \(y=k\). This Year 12 Mathematics Advanced topic (MAV-12-01) covers solving \(\sin\), \(\cos\) and \(\tan\) equations and counting solutions, in radians.
To solve \(f(x)=k\) on a given interval, find every \(x\) in that interval making the equation true. Drawing \(y=f(x)\) and the horizontal line \(y=k\), the solutions are the \(x\)-values where they cross.
Find the first solution, then use the symmetry and period of the function to reach the others — keeping only those inside the domain. \(\sin\) and \(\cos\) repeat every \(2\pi\); \(\tan\) repeats every \(\pi\).
The number of crossings in the interval is the number of solutions.
Useful base solutions (then add the period or reflect):
Method
- Sketch \(y=f(x)\) and the line \(y=k\) over the domain.
- Find the first solution from the base angle.
- Generate the rest using symmetry and the period, then keep only those in the domain.
| \(x\) | \(=\) | \(\dfrac{\pi}{6}\) |
| \(\text{or } x\) | \(=\) | \(\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}\) |
\(x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}\).
| \(x\) | \(=\) | \(\dfrac{\pi}{6}\) |
| \(\text{or } x\) | \(=\) | \(2\pi-\dfrac{\pi}{6}=\dfrac{11\pi}{6}\) |
\(x=\dfrac{\pi}{6},\ \dfrac{11\pi}{6}\).
| \(x\) | \(=\) | \(\dfrac{\pi}{4}\) |
| \(\text{or } x\) | \(=\) | \(\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}\) |
\(x=\dfrac{\pi}{4},\ \dfrac{5\pi}{4}\).
\(2x\) covers \([0,4\pi]\) — two cycles.
| \(\text{number}\) | \(=\) | \(4\) |
| \(2x\) | \(=\) | \(\dfrac{\pi}{6}\Rightarrow x=\dfrac{\pi}{12}\) |
Common pitfalls
Frequently asked questions
How do you solve a trig equation in a given domain?
Find the first solution from the base angle, then use the symmetry and period of the function to get the other solutions, and keep only those that lie inside the stated interval. Sine and cosine repeat every 2 pi and tangent every pi.
How many solutions does sin 2x = k have between 0 and 2 pi?
Because 2x runs from 0 to 4 pi, the graph completes two full cycles, so an equation like sin 2x = a half has up to four solutions in 0 to 2 pi.
What does solving graphically mean?
Draw y = f(x) and the horizontal line y = k on the same axes. Every point where the curve crosses the line is a solution, and the number of crossings is the number of solutions.
What is the period of tan x?
The period of tan x is pi, not 2 pi, so its solutions repeat every pi.