General Solution of Trig Equations
Master the general solution of trigonometric equations for Victorian Year 11 Mathematical Methods (VCAA) — capturing every angle that satisfies a sine, cosine or tangent equation.
You will learn to solve equations using exact values and reference angles, write the full family of solutions using whole-number multiples, and list the solutions that fall within a given interval.
Theory
In Year 11 Mathematical Methods (Unit 2), a trig equation such as \(\cos(x)=\tfrac12\) has infinitely many solutions because sine, cosine and tangent are periodic. The general solution is a formula, using an integer \(k\in\mathbb{Z}\), that captures every one of them at once. This page shows how to find the reference angle, apply the general-solution formula for \(\sin\), \(\cos\) and \(\tan\), handle negatives and multiple angles, and count solutions in a given interval — all in radians with exact values.
Because \(y=\sin x\) and \(y=\cos x\) repeat every \(2\pi\), and \(y=\tan x\) repeats every \(\pi\), a horizontal line \(y=c\) cuts each graph again and again. Every crossing is a solution, so a trig equation has infinitely many solutions.
The general solution packages all of those solutions into a single formula built from an integer parameter \(k\in\mathbb{Z}\). Each whole-number value of \(k\) selects one solution; running \(k\) through \(\dots,-2,-1,0,1,2,\dots\) generates them all.
The starting point is the reference angle \(\alpha\): the acute angle whose trig ratio equals the positive value, found with the inverse function, e.g. \(\alpha=\cos^{-1}\!\left(\tfrac12\right)=\dfrac{\pi}{3}\). The sign of the original value then decides which quadrants the solutions sit in.
Let \(\alpha\) be the reference angle and \(k\in\mathbb{Z}\). The three general-solution families are:
Take \(\alpha\) from the positive value. If \(c\) is negative, keep the same \(\alpha\) but shift to the correct quadrants: for \(\cos(x)=-c\) use \(\pi-\alpha\) in place of \(\alpha\); for \(\sin(x)=-c\) the solutions are \(2k\pi+(\pi+\alpha)\) and \(2k\pi+(2\pi-\alpha)\).
How to find the general solution
- Isolate the ratio: rearrange to \(\sin(\theta)=c\), \(\cos(\theta)=c\) or \(\tan(\theta)=c\), where \(\theta\) may be \(x\), \(2x\), \(\tfrac{\pi t}{6}\), etc.
- Reference angle: \(\alpha=\) inverse function of the positive value, as an exact multiple of \(\pi\).
- Choose the family: apply the \(\cos\), \(\sin\) or \(\tan\) formula, adjusting the quadrants if \(c\) is negative.
- Solve for \(x\): if the angle was \(2x\) or \(\tfrac{\pi t}{6}\), divide or multiply every term (including \(2k\pi\)) to make \(x\) or \(t\) the subject.
- Count if asked: substitute \(k=\dots,-1,0,1,\dots\) and keep only the values inside the required interval.
Reference angle — inverse cosine of the positive value:
| \(\alpha\) | \(=\) | \(\cos^{-1}\!\left(\dfrac{1}{2}\right)\) |
| \(=\) | \(\dfrac{\pi}{3}\) |
Cosine is positive, so use the cosine family \(x=2k\pi\pm\alpha\).
Apply the cosine general solution:
| \(x\) | \(=\) | \(2k\pi\pm\alpha\) |
| \(x\) | \(=\) | \(2k\pi\pm\dfrac{\pi}{3},\ k\in\mathbb{Z}\) |
General solution: \(x=2k\pi\pm\dfrac{\pi}{3},\ k\in\mathbb{Z}\).
Reference angle:
| \(\alpha\) | \(=\) | \(\sin^{-1}\!\left(\dfrac{1}{2}\right)\) |
| \(=\) | \(\dfrac{\pi}{6}\) |
Sine is positive, so solutions lie in the first and second quadrants — two families.
Apply the sine general solution:
| \(x\) | \(=\) | \(2k\pi+\alpha\ \text{ or }\ 2k\pi+(\pi-\alpha)\) |
| \(x\) | \(=\) | \(2k\pi+\dfrac{\pi}{6}\ \text{ or }\ 2k\pi+\dfrac{5\pi}{6}\) |
General solution: \(x=2k\pi+\dfrac{\pi}{6}\) or \(x=2k\pi+\dfrac{5\pi}{6},\ k\in\mathbb{Z}\).
Reference angle:
| \(\alpha\) | \(=\) | \(\tan^{-1}\!\left(\sqrt{3}\right)\) |
| \(=\) | \(\dfrac{\pi}{3}\) |
Tangent repeats every \(\pi\), so add integer multiples of \(\pi\) to \(\alpha\).
Apply the tangent general solution:
| \(x\) | \(=\) | \(k\pi+\alpha\) |
| \(x\) | \(=\) | \(k\pi+\dfrac{\pi}{3},\ k\in\mathbb{Z}\) |
General solution: \(x=k\pi+\dfrac{\pi}{3},\ k\in\mathbb{Z}\).
Reference angle for the sine value:
| \(\alpha\) | \(=\) | \(\sin^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{\pi}{3}\) |
General solution for the angle \(2x\) (two sine families):
| \(2x\) | \(=\) | \(2k\pi+\dfrac{\pi}{3}\ \text{ or }\ 2k\pi+\dfrac{2\pi}{3}\) |
Divide every term by \(2\) to solve for \(x\):
| \(x\) | \(=\) | \(k\pi+\dfrac{\pi}{6}\ \text{ or }\ k\pi+\dfrac{\pi}{3},\ k\in\mathbb{Z}\) |
General solution: \(x=k\pi+\dfrac{\pi}{6}\) or \(x=k\pi+\dfrac{\pi}{3},\ k\in\mathbb{Z}\).
Common pitfalls
Frequently asked questions
What is the general solution of a trig equation?
A formula, using an integer \(k\in\mathbb{Z}\), that lists every solution at once. Each value of \(k\) gives one solution; all integers together give them all.
Why are there infinitely many solutions?
Sine and cosine repeat every \(2\pi\) and tangent every \(\pi\), so the line \(y=c\) keeps cutting the graph — each crossing is another solution.
What are the general-solution formulas?
With reference angle \(\alpha\): \(\cos(x)=c\Rightarrow x=2k\pi\pm\alpha\); \(\sin(x)=c\Rightarrow x=2k\pi+\alpha\) or \(2k\pi+(\pi-\alpha)\); \(\tan(x)=c\Rightarrow x=k\pi+\alpha\).
How do I handle a negative value like \(\cos(x)=-\tfrac12\)?
Find \(\alpha\) from the positive value (\(\tfrac{\pi}{3}\)), then use the second-quadrant angle \(\pi-\alpha=\tfrac{2\pi}{3}\): \(x=2k\pi\pm\tfrac{2\pi}{3}\).
How do I solve when the angle is \(2x\) or \(3x\)?
Write the general solution for the whole angle first, then divide every term by \(2\) (or \(3\)) to make \(x\) the subject.
What is a reference angle?
The acute angle \(\alpha\) whose trig ratio equals the positive value, e.g. \(\alpha=\sin^{-1}\!\left(\tfrac12\right)=\tfrac{\pi}{6}\).