Precalculus

Trigonometry

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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

First-Order Homogeneous Trigonometric Equation

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

Second-Order Homogeneous Trigonometric Equation

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

Third-Order Homogeneous Trigonometric Equation

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

Click or tap a problem to see the solution.

Example 1

Solve the equation

\[\sin x + \cos x = 0.\]

Example 2

Find the general solution of the equation

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

Example 3

Solve the trigonometric equation

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

Example 4

Solve the equation

\[3\sin 2x - \cos 2x = 3.\]

Example 5

Solve the trigonometric equation

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

Example 6

Find the general solution of the equation

\[\sin x + \cos x = \frac{1}{{\cos x}}.\]

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:

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

Answer:

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

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

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