Trigonometric equations with domain restrictions

2026-10-22 huynhvietkhanh Comments: 0

Retain denominator and square-root restrictions, filter periodic solution families, and reject extraneous roots introduced by squaring.

High schoolTrigonometry

Content updated: October 22, 2026

Simplification may remove a denominator from the written expression, but it does not enlarge the domain of the original equation. The final solution set is the intersection of the candidates with every restriction stated before transforming the equation.

Learning goals: find the domain, clear denominators under the correct conditions, exclude angle classes, and check solutions after squaring.

Prerequisites: basic trigonometric equations, double-angle formulas, factorization, and square roots.

On this page: The idea · Method · Examples · Practice · Further learning

1. Retain the domain throughout the solution

Tangent requires cosine different from $0$; cotangent requires sine different from $0$; a fraction requires every denominator different from $0$. For a real square root, the radicand must be nonnegative. If the root occurs in a denominator, the radicand must be positive. Take the intersection of all simultaneous restrictions.

Multiplying both sides by a denominator known to differ from $0$ is an equivalent transformation on that domain. Cancelling a factor also requires that factor to differ from $0$. Squaring both sides gives only an implication unless signs have been checked; solutions found after squaring may need direct verification.

2. Filtering solution families

Write forbidden angles with the same period as the candidate solution families. If a family has step $2\pi/q$ with $q$ a positive integer, write its index as $k=qn+r$, where $r=0,\ldots,q-1$. Each value of $r$ determines an angle class modulo $2\pi$; compare it with the forbidden angles.

These classes can also be shown on the unit circle. A diagram supports the comparison, but the solution must state which classes are excluded and why, so it does not depend on an unseen figure.

3. Worked examples

Example 1. A trigonometric denominator

Solve $\dfrac{\cos x-\sin2x}{2\cos^2x-\sin x-1}=\sqrt3$. Set $s=\sin x$. The denominator is $1-s-2s^2=(1+s)(1-2s)$, so require $\sin x\ne-1$ and $\sin x\ne1/2$. The forbidden angles modulo $2\pi$ are $-\pi/2,\pi/6,5\pi/6$.

On this domain, clearing the denominator and rearranging yields $\cos x+\sqrt3\sin x=\sqrt3\cos2x+\sin2x$, equivalently $\sin(x+\pi/6)=\sin(2x+\pi/3)$. The two candidate families are

$$x=-\frac\pi6+2k\pi\quad\text{or}\quad x=\frac\pi6+\frac{2k\pi}{3}.$$

The first family has sine $-1/2$, different from both forbidden values, so it is accepted. For the second family, write $k=3n+r$. For $r=0,1,2$, the corresponding angles modulo $2\pi$ are $\pi/6,5\pi/6,3\pi/2$; all three are forbidden. Thus the final solutions are only $x=-\pi/6+2k\pi$.

Checking at $x=-\pi/6$: the numerator is $\sqrt3$ and the denominator is $1$, so the fraction equals $\sqrt3$ as required. Alternatively, factor the numerator as $\cos x(1-2\sin x)$ and denominator as $(1+\sin x)(1-2\sin x)$, retaining both original restrictions.

Example 2. Cancellation does not restore forbidden points

Solve $(1-\cos2x)/\sin x=0$. Require $\sin x\ne0$. On this domain, the fraction equals $2\sin x$; if it equals $0$, sine must equal $0$, contrary to the restriction. There are no solutions, even though the simplified expression itself is defined at the forbidden points.

Example 3. Squaring and checking signs

Solve $\sqrt{\sin x}=\sin x$. The restriction is $s=\sin x\in[0;1]$. Both sides are nonnegative, so squaring is equivalent to $s=s^2$. Accept $s=0$ or $s=1$: the solutions are $x=k\pi$ or $x=\pi/2+2k\pi$. Every solution satisfies both the restriction and the original equation.

4. Practice exercises

Solve $\tan x=\sin x/\cos x$.
Solve $\sin x/(1-\cos x)=0$.
Solve $\sqrt{\sin x}=-\sin x$.

Show hints and answers

The sides agree on the domain $\cos x\ne0$. Thus every $x\in\mathbb R\setminus\{\pi/2+k\pi:k\in\mathbb Z\}$ is a solution.
Require $\cos x\ne1$. The numerator equals $0$ when $x=k\pi$; only odd multiples of $\pi$ have denominator different from $0$. The solutions are $x=\pi+2k\pi$.
Require $\sin x\ge0$; the left-hand side is nonnegative and the right-hand side is nonpositive, so they can be equal only when $\sin x=0$. Hence $x=k\pi$. Squaring alone would also admit the extraneous value sin x = 1.

5. Common mistakes

Retain restrictions after cancellation. Do not multiply by a denominator before determining whether it equals $0$ or differs from $0$. Do not reject an entire family from only a few index values; examine all residue classes. After squaring, retain the sign conditions or substitute directly into the original equation.

6. Related lessons

Original source resources

These original source resources have not been translated.

Original reference figure 1