Precalculus

Trigonometry

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Basic Trigonometric Equations

Angles (arguments of functions): x, x1, x2
Set of integers: Ζ
Integer: n
Real number: a

Trigonometric functions: sin x, cos x, tan x, cot x
Inverse trigonometric functions: arcsin a, arccos a, arctan a, arccot a

An equation involving trigonometric functions of an unknown angle is called a trigonometric equation.

Basic trigonometric equations have the form

\[\sin x = a,\;\cos x = a,\;\tan x = a,\;\cot x = a,\]

here \(x\) is an unknown, \(a\) is any real number.

Equation \(\sin x = a\)

If \(\left| a \right| \gt 1\), the equation \(\sin x = a\) has no solutions.

If \(\left| a \right| \le 1,\) the general solution of the equation \(\sin x = a\) is written as

\[x = \left({-1}\right)^n \arcsin a + \pi n,\;n \in \mathbb{Z}.\]

This formula contains two branches of solutions:

\[{x_1} = \arcsin a + 2\pi n,\;{x_2} = \pi - \arcsin a + 2\pi n,\;n \in \mathbb{Z}.\]
Solution of the equation sin(x)=a
Figure 1.

The solutions of a trigonometric equation that lie in the interval \(\left[ {0,2\pi } \right)\) are called principal solutions. The principal solutions for the equation \(\sin x = a\) are

\[\arcsin a,\;\pi - \arcsin a.\]

In the simple case \(\sin x = 1\) the general solution has the form

\[x = \pi/2 + 2\pi n,\;n \in \mathbb{Z}.\]

Similarly, the solution of the equation \(\sin x = -1\) is given by

\[x = -\pi/2 + 2\pi n,\; n \in \mathbb{Z}.\]

Case \(\sin x = 0\) (zeroes of the sine):

\[x = \pi n,\; n \in \mathbb{Z}.\]

Equation \(\cos x = a\)

If \(\left| a \right| \gt 1,\) the equation \(\cos x = a\) has no solutions.

If \(\left| a \right| \le 1,\) the general solution of the equation \(\cos x = a\) has the form

\[x = \pm\arccos a + 2\pi n,\;n \in \mathbb{Z}.\]

This formula includes two sets of solutions:

\[{x_1} = \arccos a + 2\pi n,\; {x_2} = -\arccos a + 2\pi n,\; n \in \mathbb{Z}.\]
Solution of the equation cos(x)=a
Figure 2.

In the case \(\cos x = 1\), the solution is written as

\[x = 2\pi n,\; n \in \mathbb{Z}.\]

Case \(\cos x = -1:\)

\[x = \pi + 2\pi n,\; n \in \mathbb{Z}.\]

Case \(\cos x = 0\) (zeroes of the cosine):

\[x = \pi/2 + \pi n,\; n \in \mathbb{Z}.\]

Equation \(\tan x = a\)

For any value of \(a\), the general solution of the equation \(\tan x = a\) has the form

\[x = \arctan a + \pi n,\;n \in \mathbb{Z}.\]
Solution of the equation tan(x)=a
Figure 3.

Case \(\tan x = 0\) (zeroes of the tangent):

\[x = \pi n,\; n \in \mathbb{Z}.\]

Equation \(\cot x = a\)

For any value of \(a\), the general solution of the trigonometric equation \(\cot x = a\) is written as

\[x = \text{arccot}\, a + \pi n,\;n \in \mathbb{Z}.\]
Solution of the equation cot(x)=a
Figure 4.

Case \(\cot x = 0\) (zeroes of the cotangent):

\[x = \pi/2 + \pi n,\; n \in \mathbb{Z}.\]

Solved Problems

Example 1.

Solve the equation

\[\sin x = - \frac{1}{2}.\]

Solution.

The general solution of the equation has the form

\[x = {\left( { - 1} \right)^n}\arcsin \left( { - \frac{1}{2}} \right) + \pi n,\;n \in \mathbb{Z}.\]

The inverse sine function is odd, so that

\[\arcsin \left( { - \frac{1}{2}} \right) = - \arcsin \frac{1}{2} = - \frac{\pi }{6}.\]

Hence, the general solution is written as

\[x = {\left( { - 1} \right)^n}\left( { - \frac{\pi }{6}} \right) + \pi n = {\left( { - 1} \right)^{n + 1}}\frac{\pi }{6} + \pi n,\;n \in \mathbb{Z}.\]

The principal solutions on the interval \(\left[ {0,2\pi } \right)\) are given by

\[\begin{array}{*{20}{l}} {n = 1:}&{x_1} = \frac{\pi }{6} + \pi = \frac{{7\pi }}{6}\\ {n = 2:}&{x_2} = - \frac{\pi }{6} + 2\pi = \frac{{11\pi }}{6} \end{array}\]

Example 2.

Solve the equation

\[\cos \left( {x + \frac{\pi }{3}} \right) = - 1.\]

Solution.

In this special case, the general solution is given by

\[x + \frac{\pi }{3} = \pi + 2\pi n,\;n \in \mathbb{Z}.\]

Solve it for \(x:\)

\[x = \pi - \frac{\pi }{3} + 2\pi n = \frac{{2\pi }}{3} + 2\pi n,\;n \in \mathbb{Z}.\]

The principal solution contains one value:

\[n = 0:\;{x_0} = \frac{{2\pi }}{3}.\]

Example 3.

Find the general solution of the equation

\[\sqrt 3 \sin x = \cos x.\]

Solution.

We rewrite the equation in the form

\[\sqrt 3 \sin x - \cos x = 0\]

and divide both sides by \(\cos x.\) Note that \(\cos x \ne 0.\) Indeed, if \(\cos x = 0,\) then

\[\sin x = \pm \sqrt {1 - {{\cos }^2}x} = \pm 1 \ne 0,\]

that is, \(\cos x = 0\) cannot be a solution of the equation. So we have

\[\frac{{\sqrt 3 \sin x}}{{\cos x}} - \frac{{\cancel{{\cos x}}}}{{\cancel{{\cos x}}}} = 0, \Rightarrow \sqrt 3 \tan x - 1 = 0, \Rightarrow \tan x = \frac{1}{{\sqrt 3 }}.\]

The general solution is given by

\[x = \arctan \frac{1}{{\sqrt 3 }} + \pi n = \frac{\pi }{6} + \pi n,\;n \in \mathbb{Z}.\]

Example 4.

Find the principal solutions of the equation

\[\cot \left( {2x + \frac{\pi }{4}} \right) = - 1.\]

Solution.

First we find the general solution. Given that \(\text{arccot}\left( { - a} \right) = \pi - \text{arccot } a,\) we have

\[2x + \frac{\pi }{4} = \text{arccot} \left( { - 1} \right) + \pi n, \Rightarrow 2x + \frac{\pi }{4} = \pi - \text{arccot } 1 + \pi n, \Rightarrow 2x + \frac{\pi }{4} = \pi - \frac{\pi }{4} + \pi n, \Rightarrow 2x = \frac{\pi }{2} + \pi n, \Rightarrow x = \frac{\pi }{4} + \frac{{\pi n}}{2},\]

where \(n \in \mathbb{Z}.\)

The principal values lie in the interval \(\left[ {0,2\pi } \right).\) Hence, our principal solutions will be

\[\begin{array}{*{20}{l}} {n = 0:}&{{x_0} = \frac{\pi }{4}}\\ {n = 1:}&{{x_1} = \frac{\pi }{4} + \frac{\pi }{2} = \frac{{3\pi }}{4}}\\ {n = 2:}&{{x_2} = \frac{\pi }{4} + \pi = \frac{{5\pi }}{4}}\\ {n = 3:}&{{x_3} = \frac{\pi }{4} + \frac{3\pi }{2} = \frac{{7\pi }}{4}} \end{array}\]

Example 5.

Solve the equation

\[{\cos ^2}x = \frac{1}{2}.\]

Solution.

This equation has two solutions:

\[\cos x = \frac{1}{2}, \Rightarrow {x_1} = \pm \arccos \frac{1}{2} + 2\pi n,\;n \in \mathbb{Z}.\]
\[\cos x = - \frac{1}{2}, \Rightarrow {x_2} = \pm \arccos \left( { - \frac{1}{2}} \right) + 2\pi k,\;k \in \mathbb{Z}.\]

Substitute the values of inverse cosine:

\[\arccos \frac{1}{2} = \frac{\pi }{3},\;\;\arccos \left( { - \frac{1}{2}} \right) = \pi - \arccos \frac{1}{2} = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3}.\]

So the general solution is given by

\[{x_1} = \pm \frac{\pi }{3} + 2\pi n,\;\;{x_2} = \pm \frac{{2\pi }}{3} + 2\pi k,\]

where \(n, k \in \mathbb{Z}.\)

Respectively, the principal solutions of the equation are

\[x = \frac{\pi }{3},\frac{{2\pi }}{3},\frac{{4\pi }}{3},\frac{{5\pi }}{3}.\]

Example 6.

Solve the equation

\[\tan x = \cot x.\]

Solution.

We rewrite the equation as follows:

\[\tan x = \cot x, \Rightarrow \tan - \frac{1}{{\tan x}} = 0, \Rightarrow \frac{{{{\tan }^2}x - 1}}{{\tan x}} = 0, \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{{\tan }^2}x = 1}\\ {\tan x \ne 0} \end{array}} \right., \Rightarrow \tan x = \pm 1.\]

We have obtained two equations. The first equation \(\tan x = 1\) has the following solution:

\[\tan x = 1, \Rightarrow {x_1} = \arctan 1 + \pi n = \frac{\pi }{4} + \pi n,\;n \in \mathbb{Z}.\]

The second equation has a solution in the form

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

We can merge both solutions and express them with one formula:

\[x = \frac{\pi }{4} + \frac{{\pi n}}{2},\;n \in \mathbb{Z}.\]

The principal values are given by

\[x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}.\]