Quadratic Trigonometric Equations
In some cases, a trigonometric equation can be reduced or converted to a quadratic equation with respect to a trigonometric function.
These equations usually contain only trigonometric functions of one angle or they can be easily converted to one variable. In addition, equations of this type include only one trig function, or all functions can be reduced to one function .
As an example consider the following equation:
\[2\,{\cos ^2}x - 5\sin x + 1 = 0.\]
Using the Pythagorean identity , we replace \({\cos ^2}x\) with \(1 - {\sin ^2}x:\)
\[2\left( {1 - {{\sin }^2}x} \right) - 5\sin x + 1 = 0, \Rightarrow 2 - 2\,{\sin ^2}x - 5\sin x + 1 = 0, \Rightarrow - 2\,{\sin ^2}x - 5\sin x + 3 = 0,\]
or
\[2\,{\sin ^2}x + 5\sin x - 3 = 0.\]
If we denote \(\sin x = t,\) we obtain the quadratic equation
\[2{t^2} + 5t - 3 = 0.\]
Find the roots of the equation:
\[D = {5^2} - 4 \cdot 2 \cdot \left( { - 3} \right) = 49, \Rightarrow {t_{1,2}} = \frac{{ - 5 \pm \sqrt {49} }}{4} = \frac{{ - 5 \pm 7}}{4} = - 3,\frac{1}{2}.\]
Now we need to solve the basic trig equations. In the first case, when \(\sin x = {t_1} = -3,\) there is no solution. For \({t_2} = \frac{1}{2},\) we get
\[\sin x = {t_2} = \frac{1}{2}, \Rightarrow x = {\left( { - 1} \right)^n}\arcsin \frac{1}{2} + \pi n = {\left( { - 1} \right)^n}\frac{\pi }{6} + \pi n,\;n \in \mathbb{Z}.\]
Answer:
\[x = {\left( { - 1} \right)^n}\frac{\pi }{6} + \pi n,\;n \in \mathbb{Z}.\]
Solved Problems
Example 1.
Solve the trigonometric equation
\[{\tan ^2}x - 3\tan x - 4 = 0.\]
Solution.
We denote \(\tan x = t\) and immediately get a quadratic equation:
\[{t^2} - 3t - 4 = 0.\]
Find the roots of the equation:
\[D = {\left( { - 3} \right)^2} - 4 \cdot \left( { - 4} \right) = 25,\;{t_{1,2}} = \frac{{3 \pm \sqrt {25} }}{2} = - 1,4.\]
Returning back to the variable \(x,\) we get the following solutions:
\[\tan x = {t_1} = - 1, \Rightarrow x_1 = \arctan \left( { - 1} \right) + \pi n = - \arctan 1 + \pi n = - \frac{\pi }{4} + \pi n,\;n \in \mathbb{Z};\]
\[\tan x = {t_2} = 4, \Rightarrow {x_2} = \arctan 4 + \pi k,\;k \in \mathbb{Z}.\]
Answer:
\[{x_1} = - \frac{\pi }{4} + \pi n,\;{x_2} = \arctan 4 + \pi k,\;n,k \in \mathbb{Z}.\]
Example 2.
Find the principal solutions of the equation
\[{\cos ^2}x - \sin x = \cos 2x.\]
Solution.
Using the Pythagorean trig identity and double-angle formula for cosine, we rewrite the equation in the form
\[\cancel{1} - {\sin ^2}x - \sin x = \cancel{1} - 2\,{\sin ^2}x, \Rightarrow {\sin ^2}x - \sin x = 0.\]
We've got an incomplete quadratic equation which can be solved by factoring.
\[\sin x\left( {\sin x - 1} \right) = 0.\]
The \(1\text{st}\) solution are zeroes of the sine function:
\[\sin x = 0, \Rightarrow {x_1} = \pi n,\;n \in \mathbb{Z}.\]
The \(2\text{nd}\) solution is given by
\[\sin x = 1, \Rightarrow {x_2} = \frac{\pi }{2} + 2\pi k,\;k \in \mathbb{Z}.\]
The principal solutions of the equation include the following values:
\[x = 0,\frac{\pi }{2},\pi .\]
Example 3.
Solve the trigonometric equation
\[\cos 4x + 2\,{\cos ^2}x - 1 = 0.\]
Solution.
We express all functions in terms of the angle \(2x.\) By the double-angle formula for cosine,
\[\cos 4x = {\cos ^2}2x - {\sin ^2}2x.\]
Using the half-angle identity , we can replace
\[2\,{\cos ^2}x - 1 = \cos 2x.\]
Then the original equation becomes
\[{\cos ^2}2x - {\sin ^2}2x + \cos 2x = 0.\]
We now apply the Pythagorean trigonometric identity :
\[{\cos ^2}2x - \left( {1 - {{\cos }^2}2x} \right) + \cos 2x = 0, \Rightarrow 2\,{\cos ^2}2x + \cos 2x - 1 = 0.\]
Make the change \(\cos2x = t\) and solve the quadratic equation:
\[2{t^2} + t - 1 = 0, \Rightarrow {t_{1,2}} = \frac{{ - 1 \pm \sqrt 9 }}{4} = - 1,\frac{1}{2}.\]
The first value \({t_1} = -1\) gives the following solution:
\[\cos 2x = {t_1} = - 1, \Rightarrow 2x = \pi + 2\pi n, \Rightarrow {x_1} = \frac{\pi }{2} + \pi n,\;n \in \mathbb{Z}.\]
The root \({t_2} = \frac{1}{2}\) defines the other set of solutions:
\[\cos 2x = {t_2} = \frac{1}{2}, \Rightarrow 2x = \pm \arccos \frac{1}{2} + 2\pi k = \pm \frac{\pi }{3} + 2\pi k, \Rightarrow {x_2} = \pm \frac{\pi }{6} + \pi k,\;k \in \mathbb{Z}.\]
Hence, the answer is
\[{x_1} = \frac{\pi }{2} + \pi n,\;{x_2} = \pm \frac{\pi }{6} + \pi k,\;n,k \in \mathbb{Z}.\]
Example 4.
Solve the equation
\[\tan x + \cot x = 2.\]
Solution.
This equation includes tangent and cotangent functions that have points of discontinuity. Therefore, we first consider the domain of this equation:
\[\left\{ \begin{array}{l}
\cos x \ne 0\\
\sin x \ne 0
\end{array} \right., \Rightarrow \left\{ \begin{array}{l}
x \ne \frac{\pi }{2} + \pi n\\
x \ne \pi n
\end{array} \right., \Rightarrow x \ne \frac{{\pi n}}{2},\;n \in \mathbb{Z}.\]
Using simple algebra, we rewrite the equation in the form
\[\tan x + \frac{1}{{\tan x}} = 2, \Rightarrow {\tan ^2}x - 2\tan x + 1 = 0.\]
Introduce the new variable \(t = \tan x\) and solve the quadratic equation:
\[{t^2} - 2t + 1 = 0, \Rightarrow {\left( {t - 1} \right)^2} = 0, \Rightarrow t = 1.\]
Making the inverse substitution, we obtain the answer:
\[\tan x = t = 1, \Rightarrow x = \arctan 1 + \pi n = \frac{\pi }{4} + \pi n,\;n \in \mathbb{Z}.\]
Example 5.
Find the general solution of the equation
\[\cos 2x + 4\,{\sin ^3}x = 1.\]
Solution.
Using the double-angle formula , we get
\[1 - 2\,{\sin ^2}x + 4\,{\sin ^3}x = 1, \Rightarrow 4\,{\sin ^3}x - 2\,{\sin ^2}x = 0, \Rightarrow {\sin ^2}x\left( {2\sin x - 1} \right) = 0.\]
There are two solutions. The first solution is given by
\[{\sin ^2}x = 0, \Rightarrow \sin x = 0, \Rightarrow x = \pi n,\,n \in \mathbb{Z}.\]
Respectively, the second solution has the form
\[2\sin x - 1 = 0, \Rightarrow \sin x = \frac{1}{2}, \Rightarrow {x_2} = {\left( { - 1} \right)^k}\arcsin \frac{1}{2} + \pi k = {\left( { - 1} \right)^k}\frac{\pi }{6} + \pi k,\,k \in \mathbb{Z}.\]
Answer:
\[{x_1} = \pi n,\;{x_2} = {\left( { - 1} \right)^k}\frac{\pi }{6} + \pi k,\]
where \(n,k \in \mathbb{Z}.\)
Example 6.
Solve the equation
\[{\cos ^4}2x + 6\,{\cos ^2}2x = \frac{{25}}{{16}}.\]
Solution.
Let \({\cos ^2}2x = t.\) Then we get a quadratic equation:
\[{t^2} + 6{t^2} - \frac{{25}}{{16}} = 0, \Rightarrow 16{t^2} + 96t - 25 = 0.\]
Find its roots:
\[D = {96^2} - 4 \cdot 16 \cdot \left( { - 25} \right) = 10816 = {104^2}, \Rightarrow {t_{1,2}} = \frac{{ - 96 \pm 104}}{{32}} = - \frac{{25}}{4},\frac{1}{4}.\]
In the first case, we have no solutions as \({t_1} = - \frac{{25}}{4} \lt - 1.\) Consider the second value \({t_2} = \frac{1}{4}.\)
\[{\cos ^2}2x = {t_2} = \frac{1}{4}, \Rightarrow \cos 2x = \pm \frac{1}{2}.\]
We have here two basic equations:
\[\cos 2x = \frac{1}{2}, \Rightarrow 2x = \pm \arccos \frac{1}{2} + 2\pi n = \pm \frac{\pi }{3} + 2\pi n, \Rightarrow {x_1} = \pm \frac{\pi }{6} + \pi n,\;n \in \mathbb{Z}.\]
\[\cos 2x = - \frac{1}{2}, \Rightarrow 2x = \pm \arccos \left( { - \frac{1}{2}} \right) + 2\pi k = \pm \left( {\pi - \arccos \frac{1}{2}} \right) + 2\pi k = \pm \frac{{2\pi }}{3} + 2\pi k, \Rightarrow {x_2} = \pm \frac{\pi }{3} + \pi k,\;k \in \mathbb{Z}.\]
Figure 1. Note that \( - \frac{\pi }{6} + \frac{\pi }{2} = \frac{\pi }{3}.\) Therefore, we can combine both these families of solutions into one solution given by the formula
\[x = \pm \frac{\pi }{6} + \frac{{\pi n}}{2},\;n \in \mathbb{Z}.\]
Example 7.
Solve the trigonometric equation
\[{\cot ^4}2x + \frac{1}{{{{\sin }^4}2x}} = 25.\]
Solution.
Using the definition of cotangent, we rewrite the equation in the form
\[\frac{{{{\cos }^4}2x}}{{{{\sin }^4}2x}} + \frac{1}{{{{\sin }^4}2x}} = 25.\]
Multiply both sides by \({{{\sin }^4}2x}\) and rearrange the terms:
\[{\cos ^4}2x - 25\,{\sin ^4}2x + 1 = 0.\]
Apply now the Pythagorean trig identity to represent \({{{\sin }^4}2x}\) as
\[{\sin ^4}2x = {\left( {{{\sin }^2}2x} \right)^2} = {\left( {1 - {{\cos }^2}2x} \right)^2} = 1 - 2\,{\cos ^2}2x + {\cos ^4}2x.\]
So the equation becomes
\[{\cos ^4}2x - 25\left( {1 - 2\,{{\cos }^2}2x + {{\cos }^4}2x} \right) + 1 = 0, \Rightarrow 24\,{\cos ^4}2x - 50\,{\cos ^2}2x + 24 = 0,\]
or
\[12\,{\cos ^4}2x - 25\,{\cos ^2}2x + 12 = 0.\]
By changing \({\cos ^2}2x = t\) we get the following quadratic equation:
\[12{t^2} - 25t + 12 = 0.\]
It has the roots \({t_1} = \frac{3}{4},\) \({t_2} = \frac{4}{3}.\)
Determine the solutions corresponding to the \(1\text{st}\) root:
\[{\cos ^2}2x = {t_1} = \frac{3}{4}, \Rightarrow \cos 2x = \pm \frac{{\sqrt 3 }}{2}, \Rightarrow 2x = \pm \arccos \left( { \pm \frac{{\sqrt 3 }}{2}} \right) + 2\pi n,\;n \in \mathbb{Z}.\]
This expression contains two families of solutions. In the first case, we have
\[2x = \pm \arccos \frac{{\sqrt 3 }}{2} + 2\pi n = \pm \frac{\pi }{6} + 2\pi n, \Rightarrow {x_1} = \pm \frac{\pi }{{12}} + \pi n,\;n \in \mathbb{Z}.\]
The second family of solutions is given by
\[2x = \pm \arccos \left( { - \frac{{\sqrt 3 }}{2}} \right) + 2\pi k = \pm \left( {\pi - \arccos \frac{{\sqrt 3 }}{2}} \right) + 2\pi k = \pm \frac{{5\pi }}{6} + 2\pi k, \Rightarrow {x_2} = \pm \frac{{5\pi }}{{12}} + \pi k,\;k \in \mathbb{Z}.\]
Figure 2. We can combine both these sets and describe them with one formula:
\[x = \pm \frac{\pi }{{12}} + \frac{{\pi n}}{2} = \frac{\pi }{{12}}\left( {6n \pm 1} \right),\;n \in \mathbb{Z}.\]
The second root \({t_2} = \frac{4}{3}\) does not produce any solutions for \(x\) as \({t_2} \gt 1.\)
Hence, the answer is
\[x = \frac{\pi }{{12}}\left( {6n \pm 1} \right),\;n \in \mathbb{Z}.\]
Example 8.
Find the general solution of the equation
\[{\tan ^4}x + {\tan ^2}x + {\cot ^4}x + {\cot ^2}x = 4.\]
Solution.
We rewrite this equation in terms of tangent:
\[{\tan ^4}x + {\tan ^2}x + \frac{1}{{{{\tan }^4}x}} + \frac{1}{{{{\tan }^2}x}} = 4.\]
Note that
\[{\left( {{{\tan }^2}x + \frac{1}{{{{\tan }^2}x}}} \right)^2} = {\tan ^4}x + 2 \cdot {\tan ^2}x \cdot \frac{1}{{{{\tan }^2}x}} + \frac{1}{{{{\tan }^4}x}} = {\tan ^4}x + 2 + \frac{1}{{{{\tan }^4}x}}.\]
Therefore,
\[{\tan ^4}x + \frac{1}{{{{\tan }^4}x}} = {\left( {{{\tan }^2}x + \frac{1}{{{{\tan }^2}x}}} \right)^2} - 2.\]
We introduce the new variable
\[t = {\tan ^2}x + \frac{1}{{{{\tan }^2}x}}.\]
Then our equation is written in the form
\[{t^2} - 2 + t = 4, \Rightarrow {t^2} + t - 6 = 0.\]
The roots of the quadratic equation are \({t_1} = - 3,\) \({t_2} = 2.\)
It is obvious that
\[t = {\tan ^2}x + \frac{1}{{{{\tan }^2}x}} \gt 0,\]
so the first root \({t_1} = - 3\) does not produce any solutions. Consider the second root \({t_2} = 2\) and make one more replacement \({\tan ^2}x = z.\) We get the following equation for \(z:\)
\[z + \frac{1}{z} = 2, \Rightarrow \frac{{{z^2} - 2z + 1}}{z} = 0, \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{{{\left( {z - 1} \right)}^2} = 0}\\
{z \ne 0}
\end{array}} \right., \Rightarrow z = 1.\]
Thus, we have
\[{\tan ^2}x = 1, \Rightarrow \tan x = \pm 1, \Rightarrow {x_{1,2}} = \arctan \left( { \pm 1} \right) + \pi n = \pm \frac{\pi }{4} + \pi n,\;n \in \mathbb{Z}.\]
We can describe both these solutions by one formula
\[x = \frac{\pi }{4} + \frac{{\pi n}}{2},\;n \in \mathbb{Z}.\]
