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Basic trigonometric equations: complete solution families

2026-10-22 huynhvietkhanh Comments: 0

Solve sine, cosine, tangent, and cotangent equations using complete families, principal-value conventions, and interval restrictions.

High schoolTrigonometry

Content updated: October 22, 2026

A trigonometric value usually corresponds to many angles. Solving an equation therefore requires more than one calculator value: find every solution family and retain the domain restrictions.

Learning goals: solve the four basic trigonometric equations, distinguish principal values from general solutions, and select solutions in an interval.

Prerequisites: the unit circle, radians, periods, and special trigonometric values.

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

1. Domains and angle units

Throughout this lesson, $x$ is a real angle measured in radians and $k\in\mathbb Z$. If using degrees, replace $2\pi$ by $360^\circ$ and $\pi$ by $180^\circ$ consistently throughout; do not mix the two units.

Since $-1\le\sin x,\cos x\le1$, the equations $\sin x=a$ and $\cos x=a$ have no real solutions when $|a|>1$. The function $\tan x$ is defined only when $\cos x\ne0$, while $\cot x$ requires $\sin x\ne0$. On these domains, tangent and cotangent attain every real value.

2. Four general-solution formulas

$$\begin{aligned}\sin x=\sin\alpha&\Longleftrightarrow x=\alpha+2k\pi\ \text{or}\ x=\pi-\alpha+2k\pi,\\\cos x=\cos\alpha&\Longleftrightarrow x=\alpha+2k\pi\ \text{or}\ x=-\alpha+2k\pi.\end{aligned}$$

The two sine branches correspond to two points sharing a y-coordinate on the circle; the cosine branches share an x-coordinate. For values $1$ or $-1$, the branches may coincide as one family. For the value $0$, the solutions can be combined into a family with period $\pi$.

$$\begin{aligned}\tan x=\tan\alpha&\Longleftrightarrow x=\alpha+k\pi,\quad \cos\alpha\ne0,\\\cot x=\cot\alpha&\Longleftrightarrow x=\alpha+k\pi,\quad \sin\alpha\ne0.\end{aligned}$$

The conditions on the right ensure that the expression involving $\alpha$ is defined. For example, $\tan x=\tan\alpha$ implies $\sin(x-\alpha)=0$ by multiplying by two cosines different from $0$. The converse follows from period $\pi$ and preserves the domain.

For principal angles, use $\arcsin a\in[-\pi/2;\pi/2]$, $\arccos a\in[0;\pi]$, and $\arctan a\in(-\pi/2;\pi/2)$. Here $\operatorname{arccot}a$ is defined as the unique angle in $(0;\pi)$ having cotangent $a$. A principal angle is only an intermediate step toward the complete solution families.

3. Eight basic examples

Example 1. Sine of a special angle

The equation $\sin x=1/2=\sin(\pi/6)$ gives $x=\pi/6+2k\pi$ or $x=5\pi/6+2k\pi$. Writing only $x=\pi/6$ omits the second branch and all full turns.

Example 2. A nonspecial sine value

Solve $\sin x=1/3$. Set $\alpha=\arcsin(1/3)$. Then $\sin\alpha=1/3$, and the solutions are $x=\alpha+2k\pi$ or $x=\pi-\alpha+2k\pi$. No rounding of $\alpha$ is needed for an exact answer.

Example 3. Cosine of a special angle

The equation $\cos x=1/2=\cos(\pi/3)$ gives $x=\pi/3+2k\pi$ or $x=-\pi/3+2k\pi$. These families may be written compactly as $x=\pm\pi/3+2k\pi$.

Example 4. A nonspecial cosine value

Solve $\cos x=1/3$. Set $\beta=\arccos(1/3)$. The equation becomes $\cos x=\cos\beta$, so $x=\pm\beta+2k\pi$. The cosine function must be retained in this transformation.

Example 5. Tangent of a given angle

Assuming $\cos\alpha\ne0$, the equation $\tan x=\tan\alpha$ has solutions $x=\alpha+k\pi$. For example, $\alpha=\pi/4$ gives $x=\pi/4+k\pi$. The angle $\alpha$ must not be omitted from the formula.

Example 6. Tangent equal to 3

Solve $\tan x=3$. Setting $\gamma=\arctan3$ gives $x=\gamma+k\pi$. Since $\gamma\in(-\pi/2;\pi/2)$, every solution has cosine different from $0$ and belongs to the domain.

Example 7. Cotangent of a special angle

The equation $\cot x=1/\sqrt3=\cot(\pi/3)$ gives $x=\pi/3+k\pi$. Every solution has sine different from $0$.

Example 8. Cotangent equal to 3

Solve $\cot x=3$. Set $\delta=\operatorname{arccot}3$, or use $\delta=\arctan(1/3)$ since this value lies in $(0;\pi/2)$. The solutions are $x=\delta+k\pi$.

4. Practice exercises

  1. Solve $\sin x=0$ and $\cos x=-1$.
  2. Solve $\sin(2x)=1/2$ on $[0;\pi)$.
  3. Solve $\tan x=-1$, and decide whether $\sin x=3/2$ has a solution.
Show hints and answers
  1. $\sin x=0$ gives $x=k\pi$; $\cos x=-1$ gives $x=\pi+2k\pi$.
  2. From $2x=\pi/6+2k\pi$ or $2x=5\pi/6+2k\pi$, we obtain $x=\pi/12+k\pi$ or $x=5\pi/12+k\pi$. In the given interval, only $\pi/12,5\pi/12$ occur.
  3. $x=-\pi/4+k\pi$. The sine equation has no solution because $3/2>1$.

5. Common mistakes

Do not omit the reference angle in tangent or cotangent formulas. Do not use period $2\pi$ in a way that captures only some tangent or cotangent solutions. When dividing an angle expression such as $2x$, divide the period as well. Check the domain first and filter the interval after finding the general solution.

6. Related lessons

7. Resources from the original article

Original source resources

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