# Homogeneous Trigonometric Equations

A trigonometric equation is said to be a homogeneous equation in sin x and cos x if the degrees of all terms in the equation are the same. This degree is called the degree of the homogeneous equation.

For example, the equation of the form

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

is a first-order homogeneous equation. Notice that a homogeneous equation has zero on the right-hand side of the equality sign.

A second-order homogeneous trigonometric equation is written as

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

A similar second-order equation of the form

$a\,{\sin ^2}x + b\sin x\cos x + c\,{\cos ^2}x = d$

is not a homogeneous equation since its right-hand side is not equal to zero. But we can reduce it to a homogeneous equation using the Pythagorean identity:

$a\,{\sin ^2}x + b\sin x\cos x + c\,{\cos ^2}x = d \cdot 1,$
$\Rightarrow a\,{\sin ^2}x + b\sin x\cos x + c\,{\cos ^2}x = d\left( {{{\sin }^2}x + {{\cos }^2}x} \right),$
$\Rightarrow \left( {a - d} \right){\sin ^2}x + b\sin x\cos x + \left( {c - d} \right){\cos ^2}x = 0.$

A third-order homogeneous trigonometric equation is given by

$a\sin^3x + b\sin^2x\cos x + c\sin x\cos^2x + d\cos^3x = 0$

and so on.

By dividing by $${\cos ^n}x,$$ a homogeneous equation of degree $$n$$ can be reduced to an algebraic equation with respect to the function $$\tan x.$$

## Solved Problems

### Example 1.

Solve the equation

$\sin x + \cos x = 0.$

Solution.

We divide both sides of the equation by $$\cos x.$$ This can be done because $$\cos x \ne 0.$$ Indeed, cosine and sine in the left-hand side cannot be simultaneously equal to zero, so $$\cos x = 0$$ is not a solution of the equation.

$\frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{\cos x}} = \frac{0}{{\cos x}}, \Rightarrow \tan x + 1 = 0, \Rightarrow \tan x = - 1.$

The basic equation $$\tan x = -1$$ has the following solution:

$x = \arctan \left( { - 1} \right) + \pi n = - \arctan 1 + \pi n = - \frac{\pi }{4} + \pi n,\,n \in \mathbb{Z}.$

### Example 2.

Find the general solution of the equation

${\sin ^2}3x = 3\,{\cos ^2}3x.$

Solution.

We deal here with a second-order homogeneous equation. Divide both sides by $${\cos ^2}3x$$ and take the square root:

${\tan ^2}3x = 3, \Rightarrow \tan 3x = \pm \sqrt 3 .$

Determine the general solution to this equation:

$3x = \arctan \left( { \pm \sqrt 3 } \right) + \pi n = \pm \frac{\pi }{3} + \pi n, \Rightarrow x = \pm \frac{\pi }{9} + \frac{{\pi n}}{3} = \frac{\pi }{9}\left( {3n \pm 1} \right),\,n \in \mathbb{Z}.$

### Example 3.

Solve the trigonometric equation

$5\,{\sin ^2}x + \sin 2x - 3\,{\cos ^2}x = 0.$

Solution.

By the double-angle formula for sine, we can write

$5\,{\sin ^2}x + 2\sin x\cos x - 3\,{\cos ^2}x = 0.$

Divide by $${\cos ^2}x:$$

$5\,{\tan ^2}x + 2\tan x - 3 = 0.$

By denoting $$\tan x = t,$$ determine the roots of the quadratic equation:

$5{t^2} + 2t - 3 = 0, \Rightarrow D = {2^2} - 4 \cdot 5 \cdot \left( { - 3} \right) = 64, \Rightarrow {t_{1,2}} = \frac{{ - 2 \pm \sqrt {64} }}{{10}} = - 1,\frac{2}{5}.$

We get two families of solutions:

$\tan x = {t_1} = - 1, \Rightarrow {x_1} = \arctan \left( { - 1} \right) + \pi n = - \frac{\pi }{4} + \pi n,\,n \in \mathbb{Z};$
$\tan x = {t_2} = \frac{2}{5}, \Rightarrow {x_2} = \arctan \frac{2}{5} + \pi k,\,k \in \mathbb{Z}.$

${x_1} = - \frac{\pi }{4} + \pi n,{x_2} = \arctan \frac{2}{5} + \pi k,\;n,k \in \mathbb{Z}.$

### Example 4.

Solve the equation

$3\sin 2x - \cos 2x = 3.$

Solution.

We have here a linear nonhomogeneous equation. However, using the double-angle formulas and Pythagorean identity, we can convert it into a second-order homogeneous equation:

$3\sin 2x - \cos 2x = 3, \Rightarrow 6\sin x\cos x - \left( {{{\cos }^2}x - {{\sin }^2}x} \right) = 3({\sin ^2}x + {\cos ^2}x), \Rightarrow 2\,{\sin ^2}x - 6\sin x\cos x + 4\,{\cos ^2}x = 0,$

or

${\sin ^2}x - 3\sin x\cos x + 2\,{\cos ^2}x = 0.$

Divide both sides by $${\cos ^2}x:$$

${\tan ^2}x - 3\tan x + 2 = 0.$

We got a quadratic trigonometric equation. Making the substitution $$\tan x = t,$$ find the roots of the equation:

${t^2} - 3t + 2 = 0,\Rightarrow {t_{1,2}} = \frac{{3 \pm 1}}{2} = 1,2.$

Hence, the original equation has two sets of solutions:

$\tan x = 1, \Rightarrow {x_1} = \frac{\pi }{4} + \pi n,\,n \in \mathbb{Z};$
$\tan x = 2, \Rightarrow {x_2} = \arctan 2 + \pi k,\,k \in \mathbb{Z}.$

### Example 5.

Solve the trigonometric equation

$4\,{\sin ^4}x + 2\,{\sin ^2}x - {\cos ^2}x = 0.$

Solution.

This is not a homogeneous equation since not all terms have the same degree. To make it homogeneous, we multiply the last two terms in the left-hand side by $$1 = {\sin ^2}x + {\cos ^2}x.$$ This yields:

$4\,{\sin ^4}x + \left( {2\,{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 0,$
$4\,{\sin ^4}x + 2\,{\sin ^4}x - {\sin ^2}x\,{\cos ^2}x + 2\,{\sin ^2}x\,{\cos ^2}x - {\cos ^4}x = 0,$
$6\,{\sin ^4}x + {\sin ^2}x\,{\cos ^2}x - {\cos ^4}x = 0.$

Now we have a fourth-degree homogeneous equation and can solve it by dividing by $${\cos ^4}x:$$

$6\,{\tan ^4}x + {\tan ^2}x - 1 = 0.$

Let $${\tan ^2}x = t.$$ Then

$6{t^2} + t - 1 = 0, \Rightarrow D = {\left( { - 1} \right)^2} - 4 \cdot 6 \cdot \left( { - 1} \right) = 25, \Rightarrow {t_{1,2}} = \frac{{ - 1 \pm \sqrt {25} }}{{12}} = - \frac{1}{2},\frac{1}{3}.$

The first root $${t_1} = - \frac{1}{2}$$ does not produce any solutions:

${\tan ^2}x = {t_1} = - \frac{1}{2}, \Rightarrow x \in \varnothing.$

The second root $${t_2} = \frac{1}{3}$$ gives us the following solution:

${\tan ^2}x = {t_2} = \frac{1}{3}, \Rightarrow \tan x = \pm \frac{1}{{\sqrt 3 }}, \Rightarrow x = \arctan \left( { \pm \frac{1}{{\sqrt 3 }}} \right) + \pi n = \pm \frac{\pi }{6} + \pi n,\,n \in \mathbb{Z}.$

### Example 6.

Find the general solution of the equation

$\sin x + \cos x = \frac{1}{{\cos x}}.$

Solution.

Note that the domain of the equation is defined by the condition $$\cos x \ne 0.$$ Let's rewrite the right-hand side of the equation:

$\frac{1}{{\cos x}} = \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{\cos x}} = \frac{{{{\sin }^2}x}}{{\cos x}} + \cos x.$

Now we have a second-order homogeneous equation:

$\sin x + \cancel{{\cos x}} = \frac{{{{\sin }^2}x}}{{\cos x}} + \cancel{{\cos x}}, \Rightarrow \sin x - \frac{{{{\sin }^2}x}}{{\cos x}} = 0, \Rightarrow \sin x\cos x - {\sin ^2}x = 0.$

Factor the left-hand side:

$\sin x\left( {\cos x - \sin x} \right) = 0.$

The first set of solutions is given by

$\sin x = 0, \Rightarrow {x_1} = \pi n,\,n \in \mathbb{Z}.$

The other set of solutions follows from the first-order homogeneous equation:

$\cos x - \sin x = 0, \Rightarrow 1 - \tan x = 0, \Rightarrow \tan x = 1, \Rightarrow {x_2} = \frac{\pi }{4} + \pi k,\,k \in \mathbb{Z}.$

Hence, the general solution is given by

${x_1} = \pi n,\;{x_2} = \frac{\pi }{4} + \pi k,\;n,k \in \mathbb{Z}.$