Trigonometric equations are mathematical expressions that involve trigonometric functions (such as sine, cosine, tangent, etc.) and are set equal to a value. The goal is to find the values of the variable (usually an angle) that satisfy the equation.
Table of Contents
For example, a simple trigonometric equation might be:
sin(x) = 0.5
Solving this equation involves finding the values of x that make the sine of x equal to 0.5. The solutions could be periodic due to the nature of trigonometric functions.
Note: To solve the trigonometric equations, we will use the information that the period of sin x and cos x is 2π and the period of tan x is π.
Trigonometric Equations Examples
As Trigonometric Equations represent the relationships between different trigonometric functions, there can be infinitely many Trigonometric Equations. Some examples of Trigonometric Equations are:
- sin(x) = 1/√2
- cos(3x) = -1/2
- 2sin(2x) – 1 = 0
- tan(2x) + 3 = 0
- 2 cos(x) + sin(2x) = 1
- 3 sin(x) – 2 cos(2x) = 1
- 2 sin(3x) + tan(x) = 0
- cot(x) + 2 cos(x) = 0
- 4 cos(2x) – 3 sin(3x) = 2
How to Solve Trigonometric Equations?
The following steps define how to solve trigonometric equations:
- Step 1: Transform the supplied trigonometric equation into a single trigonometric ratio equation (sin, cos, tan).
- Step 2: Convert the equation with many angles or submultiple angles into a simple angle using the trigonometric equation.
- Step 3: Now, write the equation as a polynomial, quadratic, or linear equation.
- Step 4: Solve the trigonometric problem in the same way you would any other equation, then calculate the trigonometric ratio.
- Step 5: The solution of the trigonometric equation is represented by the angle of the trigonometric ratio or by the value of the trigonometric ratio.
General Solutions Trigonometric Equations
The table below lists the generic solutions to the trigonometric functions defined in equations.
| Trigonometric Equations | General Solutions |
|---|---|
| sin θ = 0 | θ = nπ |
| cos θ = 0 | θ = (nπ + π/2) |
| tan θ = 0 | θ = nπ |
| sin θ = 1 | θ = (2nπ + π/2) = (4n+1)π/2 |
| cos θ = 1 | θ = 2nπ |
| sin θ = sin α | θ = nπ + (-1)nα, Where α ∈ [-π/2, π/2] |
| cos θ = cos α | θ = 2nπ ± α, Where α ∈ (0, π] |
| tan θ = tan α | θ = nπ + α, Where α ∈ (-π/2 , π/2] |
| sin 2θ = sin 2α | θ = nπ ± α |
| cos 2θ = cos 2α | θ = nπ ± α |
| tan 2θ = tan 2α | θ = nπ ± α |
If α is supposed to be the least positive number that satisfies two specified trigonometrical equations, then the general value of θ will be 2nπ + α.
Principle Solution of Trigonometric Equations
The principal solution of a trigonometric equation refers to the solution that falls within a specific interval, typically between 0° and 360° or 0 and 2π radians. This solution represents the primary or fundamental solution of the equation, and it is often used as a reference point when finding other solutions.
Proof of Solutions of Trigonometric Equations
Let us now use theorems to demonstrate these solutions i.e.,
- sin x = sin y implies x = nπ + (–1)ny, where n ∈ Z
- cos x = cos y, which implies x = 2nπ ± y, where n ∈ Z
- tan x = tan y implies x = nπ + y, where n ∈ Z
Let’s discuss these theorems in detail.
Theorem 1: If x and y are real integers, sin x = sin y implies x = nπ + (–1)ny, where n ∈ Z
Proof:
Consider the following equation: sin x = sin y. Let’s try to solve this trigonometric equation in general.
sin x = sin y
⇒ sin x – sin y = 0
⇒ sin x – sin y = 0
⇒ 2cos (x + y)/2 sin (x – y)/2 = 0
⇒ cos (x + y)/2 = 0 or sin (x – y)/2 = 0Taking the common answer from both requirements, we obtain:
x = nπ + (-1)ny, where n ∈ Z
Theorem 2: For any two real integers x and y, cos x = cos y, which implies x = 2nπ ± y, where n ∈ Z.
Proof:
Likewise, the generic solution of cos x = cos y is:
cos x – cos y = 0.
⇒ 2sin (x + y)/2 sin (y – x)/2 = 0
⇒ sin (x + y)/2 = 0 or sin (x – y)/2 = 0
⇒ (x + y)/2 = nπ or (x – y)/2 = nπTaking the common answer from both criteria yields:
x = 2nπ± y, where n ∈ Z
Theorem 3: Show that tan x = tan y implies x = nπ + y, where n ∈ Z if x and y are not odd multiples of π/2.
Proof:
Similarly, we may utilise the conversion of trigonometric equations to obtain the solution to equations involving tan x or other functions.
In other words, if tan x = tan y,
Then, sin x cos x = sin y cos y
⇒ sin x cos y – sin y cos x
⇒ sin x cos y – sin y cos x = 0
⇒ sin (x – y) = 0As a result, x – y =nπ or x = nπ + y, where n ∈ Z.
Trigonometric Equations Formulas
For solving other trigonometric equations, we use some of the conclusions and general solutions of the fundamental trigonometric equations. The following are the outcomes:
- For any two real integers, x and y, sin x = sin y means that x = nπ + (-1)n y, where n ∈ Z.
- For any two real integers, x and y, Cos x = cos y implies x = 2nπ ± y, where n ∈ Z.
- If x and y are not odd multiples of π/2, then tan x = tan y implies that x = nπ + y, where n ∈ Z.
People Also Read, Trigonometric Formula.
How to Solve Trigonometric Equations – Solved Examples
Example 1: Determine the primary solution to the trigonometric equation tan x = -√3
Solution:
We have tan x = -√3 here, and we know that tan /3 = √3. So there you have it.
tan x = -√3
⇒ tan x = – tan π/3
⇒ tan x = tan(π – π/3) Alternatively, tan x = tan(2π – π/3)
⇒ tan x = tan 2π/3 OR tan x = tan 5/3.
As a result, the primary solutions of tan x = -√3 are 2π/3 and 5π/3
The primary answers are x = 2π/3 and x = 5π/3.
Example 2: Find sin 2x – sin 4x + sin 6x = 0
Solution:
Given: sin 2x – sin 4x + sin 6x = 0.
⇒ sin 2x + sin 6x – sin 4x = 0
⇒ 2sin 4x.cos 2x – sin 4x = 0
⇒ sin 4x (2cos 2x – 1) = 0
⇒ sin 4x = 0 or cos 2x = 1/2
⇒ 4x = nπ or 2x = 2nπ ± π/3As a result, the general solution to the above trigonometric problem is as follows:
⇒ x = nπ/4 or nπ ± π/6
Example 3: Determine the primary solution to the equation sin x = 1/2.
Solution:
We already know that
sin π/6 = 1/2
sin 5π/6 = sin (π – π/6)
= sin π/6 = 1/2
As a result, the primary answers are x =π/6 and x = 5π/6.
Example 4: Determine the answer to cos x = 1/2.
Solution:
In this example, we’ll use the general solution of cos x = 1/2. Because we know that cos π/3 = 1/2, we have
cos x = 1/2
cos x = cos π/3
x = 2nπ + (π/3), where n ∈ Z —- [With Cosθ = Cosα, the generic solution is θ = 2nπ + α, where n ∈ Z]
As a result, cos x = 1/2 has a generic solution of x = 2nπ + (π/3), where n ∈ Z.
Example 5: Determine the primary solutions to the trigonometric equation sin x = 3/2.
Solution:
To obtain the primary solutions of sin x = √3/2, we know that sin π/3 = √3/2 and sin (π – π/3) = √3/2
sin π/3 = sin 2π/3 = √3/2
We can discover additional values of x such that sin x = √3/2, but we only need to find those values of x where x is between [0, 2π] since a primary solution is between 0 and 2π.
As a result, the primary solutions of sin x = √3/2 are x = π/3 and 2π/3.
How to Solve Trigonometric Equations – Practice Questions
Problem 1: Solve for x in the equation: sin(x) = 1/2
Problem 2: Find all solutions for x in the equation: 2 cos(2x) = 1
Problem 3: Determine the solutions for x in the equation: tan(x) = -√3
Problem 4: Solve for x in the equation: 3 sin(x) – 4 cos(x) = 0
Problem 5: Find the solutions for x in the equation: 2 sin(2x) + 1 = 0
Problem 6: Solve for x in the equation: cot(x) = 1
Problem 7: Determine all solutions for x in the equation: 3 sin(x) + 2 cos(x) = 0
Problem 8: Find the values of x that satisfy the equation: tan(2x) = 1
Problem 9: Solve for x in the equation: sec(x) = -2
Problem 10: Find all solutions for x in the equation: 4 sin(3x) = 1
Trigonometric Equations – FAQs
Define Trigonometric Equations.
Trigonometric equations are similar to algebraic equations in that they might be linear, quadratic, or polynomial equations. In trigonometric equations, the trigonometric ratios Sinθ, Cosθ, Tanθ are used to denote the variables.
Give Some Examples of Trigonometric Equations.
The following are some instances of trigonometric equations:
- 2 Cos2x + 3 Sinx = 0
- Cos4x = Cos2x
- Sin2x – Sin4x + Sin6x = 0
Which three trigonometric equations are there?
Sinθ, Cosθ, and Tanθ are the three major functions in trigonometry.
What is Sin Inverse?
The arcsin function is the inverse of the sin function. Sin, on the other hand, will not be invertible since it is not injective, and so it is not mixed (invertible). Furthermore, in order to obtain the arcsine function, the domain of sine must be limited to [−π/2,π/2].
How to Solve Trigonometric Equations?
To find solutions to trigonometric equations, use identities and algebraic techniques to isolate the variable. Then, apply inverse trig functions and consider the unit circle.
How do I solve Trigonometric Equations in Class 11 Math?
To solve trigonometric equations in Class 11 math:
- Isolate the trigonometric function on one side.
- Apply trigonometric identities and simplify the equation.
- Solve for the variable using inverse trigonometric functions.
- Check for extraneous solutions.
- State the solution with appropriate constraints.
What is the Principal Solution of a Trigonometric Equation?
The principal solution of a trigonometric equation is the solution within the principal interval, typically in radians: [-π, π] for cosine, or [0, 2π] for the sine and tangent function.
What Are Trigonometric Identities, and How Are They Used in Solving Equations?
Trigonometric identities are equations involving trig functions, and these are used to simplify and solve trigonometric equations by manipulating expressions.
How Do I Find All Solutions of a Trigonometric Equation?
To find all solutions of a trigonometric equation, you can first find the principal solution and then use the periodicity of trigonometric functions to add integer multiples of the period.