# Homogeneous Trigonometric Equations

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Fetched from algebrica.org post 15544; source modified 2026-03-08T21:28:29.

## What are homogeneous trigonometric equations

Homogeneous trigonometric equations are [equations](../equations.md) in which all terms involve trigonometric functions, such as [sine and cosine](../sine-and-cosine.md), [raised](../powers.md) to the same degree. A first-degree homogeneous trigonometric equation is generally represented by the form:

\\[
a \\sin x + b \\cos x = 0
\\]

Each term in the equation is of degree \\(1\\) and \\( a \\) and \\( b \\) are [real coefficients](../types-of-numbers.md). A general form of a first-degree trigonometric equation is:

\\[
a \\sin x + b \\cos x + c = 0
\\]

This equation is considered homogeneous when the constant term \\( c = 0 \\). Similar considerations apply to second-degree homogeneous trigonometric equations, which include only terms of second degree. These equations are generally written in the form:

\\[
a \\sin^2 x + b \\sin x \\cos x + c \\cos^2 x = 0
\\]

##### The definition can be generalized to equations of the third, fourth, fifth degree, and so on. In these cases, all terms must involve trigonometric [functions](../functions.md) raised to the same degree, ensuring the equation remains homogeneous.

Homogeneous equations of degree \\( n \\) can be solved by dividing each term of the equation by \\( \\cos^n x \\), thereby transforming the equation into one expressed in terms of \\( \\tan x \\).

## Example 1

Let’s solve the first-degree homogeneous trigonometric equation:

\\[
\\sin x - \\sqrt{3} \\cos x = 0
\\]

* * *

Dividing the equation by \\( \\cos x \\), we obtain:

\\[
\\frac{\\sin x}{\\cos x} - \\sqrt{3} = 0
\\]

Since the ratio of sine to cosine equals the [tangent](../tangent-and-cotangent.md), we can rewrite the equation in terms of \\( \\tan x. \\)

\\[
\\tan x = \\sqrt{3}
\\]

Solving the equation we get the value:

\\[
x = \\arctan(\\sqrt{3}) = \\frac{\\pi}{3}
\\]

Since the tangent function is periodic with period \\( \\pi \\), the general solution is:

\\[
x = \\frac{\\pi}{3} + k\\pi, \\quad k \\in \\mathbb{Z}
\\]

Whenever [trigonometric identities](../trigonometric-identities.md) allow us to rewrite a trigonometric equation in homogeneous form, it is generally preferable to do so, as it can simplify the calculations and make the solution process more manageable.

## Example 2

Let us now consider the second-degree trigonometric equation:

\\[
\\sin^2 x + \\sin 2x - \\cos^2 x = 0
\\]

At first glance, this is not a homogeneous equation, because not all terms are of the same degree. In particular, \\( \\sin 2x \\) is a first-degree term, while \\( \\sin^2 x \\) and \\( \\cos^2 x \\) involve second-degree expressions. In an equation like this, we cannot proceed by dividing all terms by \\( \\cos^2 x \\), because the first-degree term does not transform into a useful expression for solving the equation. In fact, this method works only when all terms are of the same degree, as in homogeneous equations.

* * *

However, using [trigonometric identities](../trigonometric-identities.md), specifically the double angle formula, we can rewrite \\( \\sin(2x) \\) as \\( 2\\sin(x)\\cos(x) \\). The equation then becomes:

\\[
\\sin^2 x + 2\\sin x \\cos x - \\cos^2 x = 0
\\]

In this way, we have transformed the original equation into a second-degree homogeneous trigonometric equation, which can now be solved by dividing all terms by \\( \\cos^2 x \\). The equation becomes:

\\[
\\frac{\\sin^2 x}{\\cos^2 x} + \\frac{2\\sin x \\cos x}{\\cos^2 x} - \\frac{\\cos^2 x}{\\cos^2 x} = 0
\\]

* * *

Simplifying each term, we obtain:

\\[
\\tan^2 x + 2\\tan x - 1 = 0
\\]

This is a [quadratic equation](../quadratic-equations.md) in \\( \\tan x \\), and can now be solved using the [quadratic formula](../quadratic-formula.md). Let’s substitute \\( t = \\tan x \\). The equation becomes:

\\[
t^2 + 2t - 1 = 0
\\]

Using the quadratic formula, we have:

\\[
t = \\frac{-2 \\pm \\sqrt{4 + 4}}{2} = \\frac{-2 \\pm 2\\sqrt{2}}{2} = -1 \\pm \\sqrt{2}
\\]

The two solutions are:

\\[
t\_1 = -1 + \\sqrt{2}, \\quad t\_2 = -1 - \\sqrt{2}
\\]

* * *

Substituting back \\( t = \\tan x \\), we obtain the solutions:

\\[
\\tan x = -1 + \\sqrt{2} \\quad \\text{and} \\quad \\tan x = -1 - \\sqrt{2}
\\]

The general solutions are:

\\[
x\_1 = \\arctan(-1 + \\sqrt{2}) + k\\pi \\quad k \\in \\mathbb{Z}
\\]



\\[
x\_2 = \\arctan(-1 - \\sqrt{2}) + k\\pi \\quad k \\in \\mathbb{Z}
\\]