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Mathematics · Ch 3 — Trigonometry

Trigonometric equations

3.6

Trigonometric equations

A trigonometric equation is an equation in which the unknown appears only inside a trigonometric ratio — for instance sin⁡θ=12\sin\theta=\tfrac12 or 2cos⁡2x−7cos⁡x+3=02\cos^2x-7\cos x+3=0. This is a fundamentally different object from a trigonometric identity (such as sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1): an identity holds for every admissible value of the angle, while an equation holds only for particular values, which is exactly what we are asked to find.

Because every trigonometric ratio repeats itself after a fixed period (2π2\pi for sine and cosine, π\pi for tangent and cotangent), a trigonometric equation essentially never has just one answer. If one value θ0\theta_0 satisfies sin⁡θ=k\sin\theta=k, so does θ0+2π\theta_0+2\pi, θ0+4π\theta_0+4\pi, θ0−2π,…\theta_0-2\pi,\ldots — the whole family generated by the period. A trigonometric equation can also have no solution at all: since sin⁡θ\sin\theta and cos⁡θ\cos\theta never leave [−1,1][-1,1], an equation like sin⁡θ=32\sin\theta=\tfrac32 is simply unsatisfiable, no matter how far we search.

Note

There is no universal recipe for solving a trigonometric equation. Depending on the equation, useful moves include factoring it into a product of simpler trigonometric expressions, rewriting everything in one function (usually via a Pythagorean or double-angle identity), or occasionally squaring both sides — a move that must always be followed by a check, since squaring can manufacture extraneous roots that do not satisfy the original equation. We record all our answers in radians unless a problem states its interval in degrees.

Principal solution vs. general solution

Because a trigonometric equation's solution set is periodic, it is useful to separate two ideas.

  • The general solution is the complete infinite family of angles satisfying the equation, expressed with an integer parameter (traditionally nn) that sweeps through Z\mathbb Z and regenerates every valid angle via the function's periodicity.
  • The principal solution is a single representative angle from that family — specifically, the one of smallest absolute value lying in [−π,π][-\pi,\pi]. (The interval [0,2π][0,2\pi] works just as well for this purpose and gives an equally valid representative; the two conventions simply parametrise the same infinite solution set starting from a different anchor point, as Example 3.43's comparison shows.) A trigonometric equation can have up to two candidates in [−π,π][-\pi,\pi] (one from each of two quadrants where the ratio takes that value); when that happens, we always keep the numerically smaller one as the principal solution.

Restricting each ratio to a fixed window is also exactly what lets us later talk about the inverse sine, cosine or tangent. The three windows used throughout this section are:

RatioPrincipal-value windowQuadrants covered
sin⁡θ\sin\theta[−π2,π2]\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]I or IV
cos⁡θ\cos\theta[0,π][0,\pi]I or II
tan⁡θ\tan\theta(−π2,π2)\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)I or IV

For a ratio expressed via a reciprocal function (cosecant, secant, cotangent), we first flip it to sine, cosine or tangent and then read the principal value off the matching window above — e.g. cosec⁡θ=−2\operatorname{cosec}\theta=-2 is read as sin⁡θ=−12\sin\theta=-\tfrac12, and cot⁡θ=3\cot\theta=\sqrt3 is read as tan⁡θ=13\tan\theta=\tfrac1{\sqrt3}.

Building the three general-solution formulas

Rather than memorising the final formulas as arbitrary rules, it helps to see why each one takes its particular shape — the shape comes directly from turning a "ratio equals ratio" equation into a product that vanishes.

For sin⁡θ=k\sin\theta=k (−1≤k≤1-1\le k\le1): let α\alpha be the numerically smallest angle with sin⁡α=k\sin\alpha=k. Writing the equation as sin⁡θ−sin⁡α=0\sin\theta-\sin\alpha=0 and applying the sum-to-product identity turns it into a product of a cosine and a sine factor, each of which can independently be set to zero. Solving those two zero-conditions separately and then recombining the two resulting families (one running over even multiples of π\pi, the other over odd multiples, shifted by ±α\pm\alpha) collapses neatly into the single alternating-sign formula

sin⁡θ=sin⁡α ⟹ θ=nπ+(−1)nα,n∈Z.(3.13)\sin\theta=\sin\alpha \ \Longrightarrow\ \theta = n\pi + (-1)^n\alpha, \qquad n\in\mathbb Z. \qquad(3.13)

The alternating (−1)n(-1)^n is what lets one formula sweep out solutions from both the quadrant where sine equals kk directly and its "mirror" quadrant — for even nn it reproduces α\alpha shifted by a full period nπ→n\pi\to effectively 2π2\pi-blocks, for odd nn it reproduces the supplementary-angle branch.

For cos⁡θ=k\cos\theta=k (−1≤k≤1-1\le k\le1): let α∈[0,π]\alpha\in[0,\pi] satisfy cos⁡α=k\cos\alpha=k. The same idea — write cos⁡θ−cos⁡α=0\cos\theta-\cos\alpha=0, convert to a product via the sum-to-product identity, and set each factor to zero — this time produces two families that recombine into

cos⁡θ=cos⁡α ⟹ θ=2nπ±α,n∈Z.\cos\theta=\cos\alpha \ \Longrightarrow\ \theta = 2n\pi \pm \alpha, \qquad n\in\mathbb Z.

Here the symmetry is a plain ±\pm rather than an alternating sign, because cosine is an even function: cos⁡(−α)=cos⁡α\cos(-\alpha)=\cos\alpha, so both +α+\alpha and −α-\alpha (equivalently 2π−α2\pi-\alpha) are always simultaneously valid.

For tan⁡θ=k\tan\theta=k (any real kk): let α∈(−π2,π2)\alpha\in\left(-\tfrac\pi2,\tfrac\pi2\right) satisfy tan⁡α=k\tan\alpha=k. Clearing denominators in sin⁡θcos⁡θ=sin⁡αcos⁡α\dfrac{\sin\theta}{\cos\theta}=\dfrac{\sin\alpha}{\cos\alpha} gives sin⁡θcos⁡α−cos⁡θsin⁡α=0\sin\theta\cos\alpha-\cos\theta\sin\alpha=0, i.e. sin⁡(θ−α)=0\sin(\theta-\alpha)=0, whose zeros are simply every integer multiple of π\pi added to α\alpha:

tan⁡θ=tan⁡α ⟹ θ=nπ+α,n∈Z.\tan\theta=\tan\alpha \ \Longrightarrow\ \theta = n\pi + \alpha, \qquad n\in\mathbb Z.

Tangent's period is only π\pi (half of sine/cosine's), which is exactly why its general solution needs just one un-alternated family instead of two interleaved ones.

The combined form acos⁡θ+bsin⁡θ=ca\cos\theta + b\sin\theta = c

Many equations mix a cosine term and a sine term of the same angle on one side of an equation set equal to a constant. The trick is to manufacture a single auxiliary angle α\alpha so that the left side collapses into one cosine (or sine) of a shifted angle:

Put a=rcos⁡αa=r\cos\alpha and b=rsin⁡αb=r\sin\alpha, where r=a2+b2r=\sqrt{a^2+b^2} (so α\alpha is simply the polar angle of the point (a,b)(a,b), and tan⁡α=ba\tan\alpha=\tfrac ba). Substituting,

acos⁡θ+bsin⁡θ=rcos⁡αcos⁡θ+rsin⁡αsin⁡θ=rcos⁡(θ−α).a\cos\theta+b\sin\theta = r\cos\alpha\cos\theta+r\sin\alpha\sin\theta = r\cos(\theta-\alpha).

The original equation becomes rcos⁡(θ−α)=cr\cos(\theta-\alpha)=c, i.e. cos⁡(θ−α)=cr=ca2+b2\cos(\theta-\alpha)=\dfrac{c}{r}=\dfrac{c}{\sqrt{a^2+b^2}} — an ordinary cos⁡(⋅)=\cos(\cdot)= constant equation in the single unknown θ−α\theta-\alpha, solvable by the cosine formula above. Writing ca2+b2=cos⁡φ\dfrac{c}{\sqrt{a^2+b^2}}=\cos\varphi for a convenient φ\varphi, the general solution is

θ=2nπ+α±φ,n∈Z.\theta = 2n\pi + \alpha \pm \varphi, \qquad n\in\mathbb Z.

Watch out

This only works when ∣ca2+b2∣≤1\left|\dfrac{c}{\sqrt{a^2+b^2}}\right|\le 1, i.e. c≤a2+b2c \le \sqrt{a^2+b^2} (taking c≥0c\ge0 WLOG) — exactly the range where a cosine can actually attain that value. If c>a2+b2c>\sqrt{a^2+b^2}, the equation acos⁡θ+bsin⁡θ=ca\cos\theta+b\sin\theta=c has no solution whatsoever, since acos⁡θ+bsin⁡θa\cos\theta+b\sin\theta itself never exceeds a2+b2\sqrt{a^2+b^2} in magnitude (this bound is exactly what is proved via the auxiliary-angle substitution in Example 3.53).

In practice one does not need to memorise α=tan⁡−1(b/a)\alpha=\tan^{-1}(b/a) symbolically — it is far quicker to just divide the whole equation by r=a2+b2r=\sqrt{a^2+b^2} and recognise the resulting coefficients as sin⁡\sin or cos⁡\cos of a familiar angle (as Examples 3.54 does with r=2r=2, recognising 32\tfrac{\sqrt3}{2} and 12\tfrac12 as sin⁡π3\sin\tfrac\pi3 and cos⁡π3\cos\tfrac\pi3).

Summary of general solutions

Trigonometric equationGeneral solution
sin⁡θ=0\sin\theta = 0θ=nπ, n∈Z\theta = n\pi,\ n\in\mathbb Z
cos⁡θ=0\cos\theta = 0θ=(2n+1)π2, n∈Z\theta = (2n+1)\dfrac{\pi}{2},\ n\in\mathbb Z
tan⁡θ=0\tan\theta = 0θ=nπ, n∈Z\theta = n\pi,\ n\in\mathbb Z
sin⁡θ=sin⁡α, α∈[−π2,π2]\sin\theta = \sin\alpha,\ \alpha\in\left[-\dfrac\pi2,\dfrac\pi2\right]θ=nπ+(−1)nα, n∈Z\theta = n\pi + (-1)^n\alpha,\ n\in\mathbb Z
cos⁡θ=cos⁡α, α∈[0,π]\cos\theta = \cos\alpha,\ \alpha\in[0,\pi]θ=2nπ±α, n∈Z\theta = 2n\pi \pm \alpha,\ n\in\mathbb Z
tan⁡θ=tan⁡α, α∈(−π2,π2)\tan\theta = \tan\alpha,\ \alpha\in\left(-\dfrac\pi2,\dfrac\pi2\right)θ=nπ+α, n∈Z\theta = n\pi + \alpha,\ n\in\mathbb Z

How the worked examples put this to use

Examples 3.42–3.55 progressively layer these formulas onto increasingly disguised equations.

  • Finding principal solutions directly (Example 3.42): each ratio is matched to a known reference angle, its sign is used to pin down which of the two allowed quadrants applies, and the window table above hands back a single numerically-smallest angle — including reciprocal-ratio cases like cosec⁡θ=−2\operatorname{cosec}\theta=-2, first flipped to sin⁡θ=−12\sin\theta=-\tfrac12.
  • Reading off a general solution once the equation is already in "ratio = ratio" form (Examples 3.43, 3.44): the matching row of the summary table is applied immediately once a reference angle α\alpha in the right window has been identified. …