Homogeneous Trigonometric Equations

What are homogeneous trigonometric equations

1.Homogeneous trigonometric equations are trigonometric equations in which every term involves trigonometric functions, such as sine and cosine, raised to the same degree. A first-degree homogeneous equation has the general form:

$$a\sin x + b\cos x = 0$$

2.Each term has degree $1$, and $a$ and $b$ are real coefficients. The general first-degree trigonometric equation is:

$$a\sin x + b\cos x + c = 0$$

3.This equation is homogeneous when the constant term satisfies $c = 0$. The same idea extends to the second degree, where only second-degree terms appear:

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

4.

The definition generalises to the third, fourth, and higher degrees. In every case all terms must involve trigonometric functions raised to the same degree, so that the equation remains homogeneous.

5.A homogeneous equation of degree $n$ can be solved by dividing each term by $\cos^n x$, which transforms it into a polynomial equation in $\tan x$.

Dividing by cosine and the case where cosine vanishes

6.Dividing every term by $\cos^n x$ is legitimate only when $\cos x \neq 0$. The values $x = \frac{\pi}{2} + k\pi$, where the cosine vanishes, are excluded by the division and must be examined separately, otherwise some solutions may be lost.

7.The verification is immediate. When $\cos x = 0$ the sine equals $\pm 1$, so substituting these values into a homogeneous equation of degree $n$ leaves only the term in which $\sin x$ appears with the full degree $n$. For the second-degree equation $a\sin^2 x + b\sin x\cos x + c\cos^2 x = 0$ this term is $a\sin^2 x$, which reduces to $a$. The values $x = \frac{\pi}{2} + k\pi$ are therefore solutions precisely when $a = 0$, that is, when the term of highest degree in $\sin x$ is absent.

8.

In practice it is enough to check whether $x = \frac{\pi}{2} + k\pi$ satisfies the original equation. If it does, these values are added to the solutions obtained from the equation in $\tan x$. If it does not, the division by $\cos^n x$ loses nothing and the method applies without restriction.

Example 1

9.Consider the first-degree homogeneous trigonometric equation:

$$\sin x - \sqrt{3}\cos x = 0$$

10.Before dividing, observe that $x = \frac{\pi}{2} + k\pi$ gives $\sin x = \pm 1$ and $\cos x = 0$, so the left-hand side equals $\pm 1$, which is never zero. The values where the cosine vanishes are not solutions, and dividing by $\cos x$ is safe. This division gives:

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

11.Since the ratio of sine to cosine is the tangent, the equation can be rewritten in terms of $\tan x$:

$$\tan x = \sqrt{3}$$

12.The value that satisfies this equation is:

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

13.Since the tangent is periodic with period $\pi$, the general solution is:

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

14.Whenever trigonometric identities allow an equation to be rewritten in homogeneous form, doing so is generally preferable, since it simplifies the calculations.

Example 2

15.Consider the second-degree trigonometric equation:

$$\sin^2 x + \sin 2x - \cos^2 x = 0$$

16.This is not homogeneous as written, because the terms do not all share the same degree. In particular $\sin 2x$ is a first-degree term, while $\sin^2 x$ and $\cos^2 x$ are second-degree terms. Dividing every term by $\cos^2 x$ is therefore not yet possible, since the first-degree term would not transform into a useful expression. The method applies only when all terms have the same degree.

17.Using the double angle formula from the trigonometric identities, $\sin 2x$ can be rewritten as $2\sin x\cos x$. The equation becomes:

$$\sin^2 x + 2\sin x\cos x - \cos^2 x = 0$$

18.This is now a second-degree homogeneous equation. The values $x = \frac{\pi}{2} + k\pi$ give a left-hand side equal to $1$, so they are not solutions, and dividing every term by $\cos^2 x$ is admissible:

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

19.Simplifying each term gives a quadratic equation in $\tan x$:

$$\tan^2 x + 2\tan x - 1 = 0$$

20.Substituting $t = \tan x$, the equation becomes:

$$t^2 + 2t - 1 = 0$$

21.Applying the quadratic formula:

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

22.The two solutions are:

$$t_1 = -1 + \sqrt{2}, \quad t_2 = -1 - \sqrt{2}$$

23.Substituting back $t = \tan x$ gives:

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

24.The general solutions are therefore:

$$ \begin{align} x_1 &= \arctan(-1 + \sqrt{2}) + k\pi, \quad k \in \mathbb{Z} \\[6pt] x_2 &= \arctan(-1 - \sqrt{2}) + k\pi, \quad k \in \mathbb{Z} \end{align} $$

Example 3

25.A trigonometric equation with a nonzero constant term can often be reduced to a homogeneous one. Consider the second-degree equation:

$$2\sin^2 x - 3\sin x\cos x + 3\cos^2 x = 1$$

26.The constant on the right-hand side prevents the terms from sharing the same degree, since $1$ has degree zero. The Pythagorean identity provides the tool to restore homogeneity, because it allows the constant to be written as a second-degree expression:

$$1 = \sin^2 x + \cos^2 x$$

27.Replacing $1$ with this identity and moving every term to the left-hand side gives:

$$2\sin^2 x - 3\sin x\cos x + 3\cos^2 x - \sin^2 x - \cos^2 x = 0$$

28.Collecting the like terms produces a second-degree homogeneous equation:

$$\sin^2 x - 3\sin x\cos x + 2\cos^2 x = 0$$

29.The values $x = \frac{\pi}{2} + k\pi$ make the left-hand side equal to $1$, so they are not solutions and the division by $\cos^2 x$ is admissible. Dividing each term gives a quadratic equation in $\tan x$:

$$\tan^2 x - 3\tan x + 2 = 0$$

30.Substituting $t = \tan x$ and factoring the quadratic equation:

$$t^2 - 3t + 2 = (t - 1)(t - 2) = 0$$

31.The two roots are $t = 1$ and $t = 2$, which correspond to:

$$\tan x = 1 \quad \text{and} \quad \tan x = 2$$

32.The general solutions are therefore:

$$ \begin{align} x_1 &= \frac{\pi}{4} + k\pi, \quad k \in \mathbb{Z} \\[6pt] x_2 &= \arctan(2) + k\pi, \quad k \in \mathbb{Z} \end{align} $$

Example 4

33.The following equation shows a case in which the values where the cosine vanishes are genuine solutions, so the preliminary check is essential. Consider the second-degree homogeneous equation:

$$\sin x\cos x - \sqrt{3}\cos^2 x = 0$$

34.The term in $\sin^2 x$ is absent, which means the coefficient $a$ of the highest power of $\sin x$ is zero. According to the criterion established above, the values $x = \frac{\pi}{2} + k\pi$ should then be solutions. Substituting them confirms it: $\sin x = \pm 1$ and $\cos x = 0$ give $\pm 1 \cdot 0 - \sqrt{3} \cdot 0 = 0$. These values satisfy the equation and form the first family of solutions:

$$x = \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z}$$

35.For the remaining solutions, where $\cos x \neq 0$, dividing every term by $\cos^2 x$ gives an equation in $\tan x$:

$$\frac{\sin x\cos x}{\cos^2 x} - \sqrt{3}\frac{\cos^2 x}{\cos^2 x} = 0$$

36.Simplifying each term reduces the equation to:

$$\tan x - \sqrt{3} = 0$$

37.This gives $\tan x = \sqrt{3}$, whose general solution is:

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

38.The complete solution set combines the two families obtained:

$$ \begin{align} x_1 &= \frac{\pi}{2} + k\pi, \quad k \in \mathbb{Z} \\[6pt] x_2 &= \frac{\pi}{3} + k\pi, \quad k \in \mathbb{Z} \end{align} $$

39.The same conclusion can be reached by factoring the original equation as $\cos x(\sin x - \sqrt{3}\cos x) = 0$, which separates the two branches directly. The first factor reproduces the solutions where the cosine vanishes, the second reproduces those obtained from the equation in $\tan x$. This confirms that the branch $\cos x = 0$ is a true solution and would have been lost by dividing without the preliminary check.