Homogeneous Equations in Terms of sin x and cos x

 

Equations in which all terms have the exact same total degree are called homogeneous equations.

 

a) First-Degree Homogeneous Equations:

 

Let \( a, b \in \mathbb{R} \).

The equation \( ax + by = 0 \) is a first-degree homogeneous equation in two variables.

If we substitute \( \cos x \) for x and \( \sin x \) for y, we obtain a first-degree homogeneous trigonometric equation of the form:

 

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

 

These equations can be solved using techniques for linear equations. Alternatively, dividing both sides of the equation by \( \cos x \) yields:

 

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

\[ \Rightarrow a + b \displaystyle \frac{\sin x}{\cos x} = 0 \quad (\cos x \neq 0) \]

\[ \Rightarrow a + b \tan x = 0 \]

\[ \Rightarrow \tan x = -\displaystyle \frac{a}{b} \]

 

which allows us to find the solution set directly.

 

Example:

 

Find the solution set for the equation \( 3 \sin 3x \ – \ \sqrt{3} \cos 3x = 0 \).

Let’s divide both sides of the equation by \( \cos 3x \):

 

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

\[ \Rightarrow 3 \displaystyle \frac{\sin 3x}{\cos 3x} \ – \ \sqrt{3} = 0 \]

\[ \Rightarrow \tan 3x = \displaystyle \frac{\sqrt{3}}{3} \]

\[ \Rightarrow \tan 3x = \tan \displaystyle \frac{\pi}{6} \]

\[ \Rightarrow 3x = \displaystyle \frac{\pi}{6} + k\pi \]

\[ \Rightarrow x = \displaystyle \frac{\pi}{18} + \frac{k\pi}{3} \]

 

Therefore, the solution set is:

 

\[ S = \left\{ x \mid x = \displaystyle \frac{\pi}{18} + \frac{k\pi}{3}, \ k \in \mathbb{Z} \right\} \]

 

b) Second-Degree Homogeneous Equations:

 

Let \( a, b, c \in \mathbb{R} \).

The equation \( ax^2 + bxy + cy^2 = 0 \) is a second-degree homogeneous equation in two variables.

Substituting \( \cos x \) for x and \( \sin x \) for y gives a second-degree homogeneous trigonometric equation of the form:

 

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

 

To solve this type of equation, we divide both sides by \( \cos^2 x \):

 

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

\[ \Rightarrow a + b \displaystyle \frac{\sin x}{\cos x} + c \frac{\sin^2 x}{\cos^2 x} = 0 \quad (\cos x \neq 0) \]

\[ \Rightarrow a + b \tan x + c \tan^2 x = 0 \]

 

An alternative method for solving these equations involves using the double-angle identities:

 

\( \cos^2 x = \displaystyle \frac{1 + \cos 2x}{2} \),   \( \cos x \cdot \sin x = \displaystyle\frac{\sin 2x}{2} \),   and   \( \sin^2 x = \displaystyle \frac{1 – \cos 2x}{2} \).

 

Substituting these identities into the original expression:

 

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

 

\[ \Rightarrow a \displaystyle \frac{\frac{1 + \cos 2x}{2}} + b \frac{\sin 2x}{2} + c \frac{1 – \cos 2x}{2} = 0 \]

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

\[ \Rightarrow (a – c) \cos 2x + b \sin 2x + a + c = 0 \]

 

which transforms the expression into a standard linear trigonometric equation.

 

Example:

 

Find the solution set for the equation \( \sqrt{3} \sin^2 x \ – \ \sin 2x \ – \ \sqrt{3} \cos^2 x = 0 \)

 

First, rewrite \( \sin 2x \) using the double-angle identity, then divide both sides by \( \cos^2 x \):

 

\[ \sqrt{3} \sin^2 x \ – \ 2 \sin x \cdot \cos x \ – \ \sqrt{3} \cos^2 x = 0 \]

\[ \Rightarrow \sqrt{3} \displaystyle \frac{\sin^2 x}{\cos^2 x} \ – \ 2 \frac{\sin x}{\cos x} – \sqrt{3} = 0 \]

\[ \Rightarrow \sqrt{3} \tan^2 x \ – \ 2 \tan x \ – \ \sqrt{3} = 0 \]

 

Using substitution, let \( \tan x = t \):

 

\[ \Rightarrow \sqrt{3} t^2 \ – \ 2t – \sqrt{3} = 0 \]

\[ \Rightarrow (\sqrt{3} t + 1)(t \ – \ \sqrt{3}) = 0 \]

\[ \Rightarrow t = -\displaystyle \frac{1}{\sqrt{3}} \quad \text{or} \quad t = \sqrt{3} \]

 

Substituting back \( \tan x \):

 

\[ \Rightarrow \tan x = -\displaystyle \frac{1}{\sqrt{3}} \quad \text{or} \quad \tan x = \sqrt{3} \]

\[ \Rightarrow \tan x = \tan \displaystyle \frac{5\pi}{6} \quad \text{or} \quad \tan x = \tan \frac{\pi}{3} \]

 

Thus, the general solutions are:

 

\[ x = \displaystyle \frac{5\pi}{6} + k\pi \quad \text{or} \quad x = \displaystyle \frac{\pi}{3} + k\pi \quad (k \in \mathbb{Z}) \]

 

The solution set is:

 

\[ S = \left\{ x \mid x = \displaystyle \frac{5\pi}{6} + k\pi \quad \text{or} \quad x = \displaystyle \frac{\pi}{3} + k\pi, \ k \in \mathbb{Z} \right\} \]

 

Example:

 

Find the solution set for the equation \( 4 \sin^2 x \ – \ \sin x \cdot \cos x + \cos^2 x = 2 \).

 

Using the Pythagorean identity, we can rewrite the constant 2 as \( 2(\sin^2 x + \cos^2 x) \) to make the equation homogeneous:

\[ 4 \sin^2 x \ – \ \sin x \cdot \cos x + \cos^2 x = 2 \]

\[ \Rightarrow 4 \sin^2 x \ – \ \sin x \cdot \cos x + \cos^2 x = 2(\sin^2 x + \cos^2 x) \]

 

Grouping like terms on one side yields:

 

\[ \Rightarrow 2 \sin^2 x \ – \ \sin x \cdot \cos x – \cos^2 x = 0 \]

 

Now, divide both sides of the equation by \( \cos^2 x \):

 

\[ \Rightarrow 2 \tan^2 x \ – \ \tan x \ – \ 1 = 0 \]

 

Let \( \tan x = t \):

 

\[ \Rightarrow 2t^2 \ – \ t \ – \ 1 = 0 \]

\[ \Rightarrow (2t + 1)(t \ – \ 1) = 0 \]

\[ \Rightarrow t = -\displaystyle \frac{1}{2} \quad \text{or} \quad t = 1 \]

\[ \Rightarrow \tan x = -\displaystyle \frac{1}{2} \quad \text{or} \quad \tan x = 1 \]

 

Let \( \alpha \) be the angle such that \( \tan \alpha = -\displaystyle \frac{1}{2} \):

 

\[ \Rightarrow \tan x = \tan \alpha \quad \text{or} \quad \tan x = \tan \displaystyle \frac{\pi}{4} \]

 

Therefore:

 

\[ x = \alpha + k\pi \quad \text{or} \quad x = \displaystyle \frac{\pi}{4} + k\pi \quad (k \in \mathbb{Z}) \]

 

The solution set is:

 

\[ S = \left\{ x \mid x = \alpha + k\pi \quad \text{or} \quad x = \displaystyle \frac{\pi}{4} + k\pi, \ k \in \mathbb{Z} \right\} \]