Solution of trigonometric equations
Theory
A trigonometric equation such as \(\sin x=k\), \(\cos x=k\) or \(\tan x=k\) is solved over a stated domain (for example \([0,2\pi]\)) by first finding the base angle \(\alpha\) from the exact values, then using the symmetry of the unit circle to list every solution. Because sine, cosine and tangent are periodic, an equation usually has several solutions; without a domain the answer is written as a general solution in radians.
A trigonometric equation asks for the angles \(x\) that make a trig ratio equal a given number \(k\). Since \(\sin\) and \(\cos\) repeat every \(2\pi\) and \(\tan\) repeats every \(\pi\), such an equation has infinitely many solutions; a domain like \(0\le x\le 2\pi\) restricts the answer to a finite list. The starting point is the base angle (or reference angle) \(\alpha\) — the acute angle with \(\sin\alpha\), \(\cos\alpha\) or \(\tan\alpha\) equal to the size \(|k|\), read from the exact-value table.
The sign of \(k\) decides which quadrants contain the solutions (the CAST rule): \(\sin\) is positive in quadrants 1 and 2, \(\cos\) in quadrants 1 and 4, and \(\tan\) in quadrants 1 and 3. Unit-circle symmetry then places each solution relative to \(\alpha\): a second-quadrant angle is \(\pi-\alpha\), a third-quadrant angle is \(\pi+\alpha\), and a fourth-quadrant angle is \(2\pi-\alpha\).
An equation of the form \(a\sin x+b=0\) is first rearranged to \(\sin x=-\dfrac{b}{a}\) and then solved as above. For a multiple angle such as \(\sin(nx)=k\), substitute \(u=nx\); as \(x\) covers \([0,2\pi]\) the new variable \(u\) covers \([0,2n\pi]\), so solve over that wider interval and divide each solution by \(n\).
With base angle \(\alpha\) (the acute angle giving \(|k|\)), the solutions in \([0,2\pi]\) for a positive \(k\) follow the unit-circle symmetry:
The four quadrant angles built from the base angle \(\alpha\):
The general solutions, valid for every integer \(n\) (radians), when no domain is given:
How to solve a trigonometric equation over a domain
- Isolate the ratio. Rearrange to \(\sin x=k\), \(\cos x=k\) or \(\tan x=k\). For a multiple angle, substitute \(u=nx\) and widen the domain: if \(x\in[0,2\pi]\) then \(u\in[0,2n\pi]\).
- Base angle. Find the acute \(\alpha\) with the ratio equal to \(|k|\) from the exact values (for example \(\cos\alpha=\dfrac12\Rightarrow\alpha=\dfrac{\pi}{3}\)).
- Choose the quadrants. Use the sign of \(k\) and CAST to decide which quadrants apply — \(\sin\!:\)Q1,Q2; \(\cos\!:\)Q1,Q4; \(\tan\!:\)Q1,Q3 for positive \(k\), the other two for negative \(k\).
- List the solutions. Write each as \(\alpha\), \(\pi-\alpha\), \(\pi+\alpha\) or \(2\pi-\alpha\) and keep those in the domain; add multiples of the period to reach every value in a widened \(u\)-interval.
- Undo the substitution. For \(u=nx\), divide every \(u\)-solution by \(n\) to recover \(x\); or, if no domain was given, state the general solution.
| \(\cos\alpha\) | \(=\) | \(\dfrac12\) |
| \(\alpha\) | \(=\) | \(\dfrac{\pi}{3}\) |
| \(x\) | \(=\) | \(\dfrac{\pi}{3}\) |
| \(x\) | \(=\) | \(2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}\) |
| \(2\sin x\) | \(=\) | \(-1\) |
| \(\sin x\) | \(=\) | \(-\dfrac12\) |
| \(\alpha\) | \(=\) | \(\dfrac{\pi}{6}\) |
| \(x\) | \(=\) | \(\pi+\dfrac{\pi}{6}=\dfrac{7\pi}{6}\) |
| \(x\) | \(=\) | \(2\pi-\dfrac{\pi}{6}=\dfrac{11\pi}{6}\) |
| \(\tan\alpha\) | \(=\) | \(\sqrt3\) |
| \(\alpha\) | \(=\) | \(\dfrac{\pi}{3}\) |
| \(x\) | \(=\) | \(\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}\) |
| \(x\) | \(=\) | \(2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}\) |
| \(\sin u\) | \(=\) | \(\dfrac{\sqrt3}{2}\) |
| \(\alpha\) | \(=\) | \(\dfrac{\pi}{3}\) |
| \(u\) | \(=\) | \(\dfrac{\pi}{3},\ \dfrac{2\pi}{3},\ \dfrac{7\pi}{3},\ \dfrac{8\pi}{3}\) |
| \(x\) | \(=\) | \(\dfrac{\pi}{6},\ \dfrac{\pi}{3},\ \dfrac{7\pi}{6},\ \dfrac{4\pi}{3}\) |
Common pitfalls
Frequently asked questions
How many solutions does sin x = k have between 0 and 2 pi?
If \(-1<k<1\) there are two solutions in \([0,2\pi]\), one in each quadrant where sine has that sign. When \(k=1,-1\) or \(0\) the two coincide into one. The same count holds for \(\cos x=k\), while \(\tan x=k\) has two solutions in \([0,2\pi]\) for every \(k\).
What is the base or reference angle?
The base angle is the acute angle \(\alpha\in\left[0,\dfrac{\pi}{2}\right]\) whose sine, cosine or tangent equals \(|k|\), read from the exact values; e.g. \(\sin\alpha=\dfrac12\) gives \(\alpha=\dfrac{\pi}{6}\). The sign of \(k\) then decides the quadrants, and symmetry places each solution relative to \(\alpha\).
How do you solve cos x = k over a domain?
Find \(\alpha\) from \(|k|\), then use cosine symmetry: for positive \(k\), \(x=\alpha\) and \(x=2\pi-\alpha\) (Q1, Q4); for negative \(k\), \(x=\pi-\alpha\) and \(x=\pi+\alpha\). Keep only the values inside the stated domain.
How do you solve sin(2x) = k on the interval 0 to 2 pi?
Let \(u=2x\). As \(x\) covers \([0,2\pi]\), \(u\) covers \([0,4\pi]\), so solve \(\sin u=k\) across that wider interval, list every solution, then divide each by \(2\). The longer interval usually yields twice as many solutions for \(x\).
How do you solve a sin x + b = 0?
Rearrange to \(\sin x=-\dfrac{b}{a}\) and solve like any \(\sin x=k\): find the base angle from the size of the number and use symmetry for the correct quadrants. The same method works for \(a\cos x+b=0\) and \(a\tan x+b=0\).
What is the general solution of a trigonometric equation?
It gives every solution with an integer \(n\): for sine \(x=n\pi+(-1)^{n}\alpha\); for cosine \(x=2n\pi\pm\alpha\); for tangent \(x=n\pi+\alpha\). Substituting \(n=0,1,2,\dots\) and the negative integers lists all solutions when no domain is set.