Homogeneous Trigonometric Equations
What are homogeneous trigonometric equations
Homogeneous trigonometric equations are equations in which all terms involve trigonometric functions, such as sine and cosine, raised 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. 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 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, 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:
\[x = \frac{\pi}{3} + k \pi , k \in \mathbb{Z}\]Whenever trigonometric identities 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:
\[sin^{2} x + sin 2 x - cos^{2} x = 0\]At first glance, this is not a homogeneous equation, because not all terms are of the same degree. In particular, $sin 2 x$ 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, specifically the double angle formula, we can rewrite $sin ( 2 x )$ 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 in $tan x$, and can now be solved using the quadratic formula. Let’s substitute $t = tan x$. The equation becomes:
\[t^{2} + 2 t - 1 = 0\]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:
\[t_{1} = - 1 + \sqrt{2} , t_{2} = - 1 - \sqrt{2}\]Substituting back $t = tan x$, we obtain the solutions:
\[tan x = - 1 + \sqrt{2} \text{and} tan x = - 1 - \sqrt{2}\]The general solutions are:
\(x_{1} = arctan ( - 1 + \sqrt{2} ) + k \pi k \in \mathbb{Z}\) \(x_{2} = arctan ( - 1 - \sqrt{2} ) + k \pi k \in \mathbb{Z}\)