Trigonometric Equations
General Solutions
A trigonometric equation involves trigonometric functions of an unknown angle. Because the functions are periodic, such equations have infinitely many solutions, so we report two things: the principal solutions (those in $[0, 2\pi)$) and the general solution (a formula covering all of them).
The general solutions of the three basic equations are standard results, where $n$ is any integer ($n \in \mathbb{Z}$):
The form of each general solution mirrors the symmetry of the curve. The cosine curve is even and symmetric about the $x$-axis, so its solutions come in $\pm$ pairs spaced by full turns. The sine curve repeats with the alternating reflection captured by $(-1)^n$. The tangent function has the shortest period, $\pi$, so its solutions are spaced just $\pi$ apart. Two special equalities are worth memorising directly:
| Equation | General solution | Reason |
|---|---|---|
| $\sin x = 0$ | $x = n\pi$ | sine vanishes at every multiple of $\pi$ |
| $\cos x = 0$ | $x = (2n + 1)\dfrac{\pi}{2}$ | cosine vanishes at odd multiples of $\dfrac{\pi}{2}$ |
| $\tan x = 0$ | $x = n\pi$ | same zeros as sine |
A reliable working method: first reduce the equation to the form $\sin x = k$, $\cos x = k$ or $\tan x = k$; find an angle $y$ whose function value is $k$ (the reference angle); fix the correct quadrant using the sign of $k$; then write the general solution from the formula above.
Deeper Insight — why three different formulas, and where the $(-1)^n$ comes from: The three general-solution forms look unrelated but each is forced by exactly how many times, and where, a horizontal line $y = k$ cuts the graph in one period. A horizontal line crosses the cosine curve at two points placed symmetrically about $0$, which is precisely what $2n\pi \pm y$ encodes. The same line cuts the sine curve at two points too, but they sit symmetrically about $\tfrac{\pi}{2}$, not $0$ — and writing that asymmetry compactly is exactly what the alternating factor $(-1)^n$ achieves, flipping the sign on every other branch. Tangent, having period $\pi$ and crossing each level just once per period, needs only the simple $n\pi + y$. The danger in this topic is mechanical: blindly applying $x = n\pi + (-1)^n y$ to a cosine equation produces wrong answers. Always identify which function you actually have before reaching for a formula, and your solutions will be both complete and correct.
- $\sin x$ is positive, so $x$ lies in Quadrants I and II.
- The reference angle with sine $\dfrac{1}{2}$ is $\dfrac{\pi}{6}$ (Q1 solution).
- The Q2 solution is $\pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$.
Answer: $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$.
- $\tan x$ is positive, so $x$ lies in Quadrants I and III.
- The reference angle is $\dfrac{\pi}{3}$ (since $\tan\dfrac{\pi}{3} = \sqrt{3}$).
- The Q3 solution is $\pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$.
Answer: $x = \dfrac{\pi}{3}$ and $x = \dfrac{4\pi}{3}$.
- Find an angle whose cosine is $-\dfrac{1}{2}$: that is $\dfrac{2\pi}{3}$ (cosine negative, Q2).
- So the equation is $\cos x = \cos\dfrac{2\pi}{3}$.
- Apply $\cos x = \cos y \Rightarrow x = 2n\pi \pm y$.
Answer: $x = 2n\pi \pm \dfrac{2\pi}{3},\ n \in \mathbb{Z}$.
- An angle with this sine is $-\dfrac{\pi}{3}$, since $\sin\left(-\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2}$.
- So $\sin x = \sin\left(-\dfrac{\pi}{3}\right)$.
- Apply $\sin x = \sin y \Rightarrow x = n\pi + (-1)^n y$.
Answer: $x = n\pi + (-1)^n\left(-\dfrac{\pi}{3}\right) = n\pi - (-1)^n\dfrac{\pi}{3},\ n \in \mathbb{Z}$.
- Replace $\cos^2 x = 1 - \sin^2 x$: $2(1 - \sin^2 x) + 3\sin x = 0$.
- $-2\sin^2 x + 3\sin x + 2 = 0$, i.e. $2\sin^2 x - 3\sin x - 2 = 0$.
- Factorise: $(2\sin x + 1)(\sin x - 2) = 0$, so $\sin x = -\dfrac{1}{2}$ or $\sin x = 2$.
- $\sin x = 2$ is impossible ($|\sin x| \le 1$), so take $\sin x = -\dfrac{1}{2} = \sin\left(-\dfrac{\pi}{6}\right)$.
- $x = n\pi + (-1)^n\left(-\dfrac{\pi}{6}\right)$.
Answer: $x = n\pi - (-1)^n\dfrac{\pi}{6},\ n \in \mathbb{Z}$.
- Write $-\cot\theta = \tan\left(\dfrac{\pi}{2} + \theta\right)$, so the right side is $\tan\left(\dfrac{\pi}{2} + x + \dfrac{\pi}{3}\right) = \tan\left(x + \dfrac{5\pi}{6}\right)$.
- The equation becomes $\tan 2x = \tan\left(x + \dfrac{5\pi}{6}\right)$.
- Apply $\tan x = \tan y \Rightarrow$ angles differ by $n\pi$: $2x = n\pi + x + \dfrac{5\pi}{6}$.
- $x = n\pi + \dfrac{5\pi}{6}$.
Answer: $x = n\pi + \dfrac{5\pi}{6},\ n \in \mathbb{Z}$.
- Periodicity means trig equations have infinitely many solutions; report principal solutions in $[0, 2\pi)$ plus a general solution.
- $\sin x = \sin y \Rightarrow x = n\pi + (-1)^n y$; the $(-1)^n$ encodes the sine curve's alternating symmetry.
- $\cos x = \cos y \Rightarrow x = 2n\pi \pm y$; the $\pm$ reflects cosine's symmetry about the origin.
- $\tan x = \tan y \Rightarrow x = n\pi + y$, the simplest form because $\tan$ has period $\pi$.
- Always identify the function first, reduce to $\sin/\cos/\tan x = k$, then apply the matching formula.
Principal Solutions
Principal solutions lie in [0, 2π) (or a stated interval). Find every angle in that range satisfying the equation rather than the full family.
Sketching the unit circle helps locate all solutions in the interval.
- Principal solutions live in the stated finite interval.
- Use the unit circle to capture every solution there.
Equations using Identities
Many equations need an identity (Pythagorean, double-angle) to reduce to a single function, then factorise and solve each factor.
Always discard solutions that make a function undefined (e.g. tan at π/2).
- Use identities to get one function, then factorise.
- Reject roots where a function is undefined.