Trigonometric Equations | MCQs, Assertion & Reason, Case Study, Formulas, FAQs, Examples, Practice Questions

February 15, 2025February 13, 2025 by Neeraj Anand, Param Anand

Trigonometric equations are the equations that involve the trigonometric functions of a variable. These equations have one or more trigonometric ratios of unknown angles. For example, cos x -sin² x = 0, is a trigonometric equation that does not satisfy all the values of x. Hence for such equations, we have to find the values of x or find the solution.

We know that sin x and cos x repeat themselves after an interval of 2π, and tan x repeats itself after an interval of π. The solutions of such trigonometry equations which lie in the interval of [0, 2π] are called principal solutions. A trigonometric equation will also have a general solution expressing all the values which would satisfy the given equation, and it is expressed in a generalized form in terms of ‘n’. The general representation of these equations comprising trigonometric ratios is;

E₁(sin x, cos x, tan x) = E₂(sin x, cos x, tan x)
Where E₁ and E₂ are rational functions.

Since sine, cosine and tangent are the major trigonometric functions, hence the solutions will be derived for the equations comprising these three ratios. However, the solutions for the other three ratios such as secant, cosecant and cotangent can be obtained with the help of those solutions.

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.

Solutions for Trigonometric Equations

Let us begin with a basic equation, sin x = 0. The principal solution for this case will be x = 0, π, 2π as these values satisfy the given equation lying in the interval [0, 2π]. But, we know that if sin x = 0, then x = 0, π, 2π, π, -2π, -6π, etc. are solutions of the given equation. Hence, the general solution for sin x = 0 will be, x = nπ, where n∈I.

Similarly, general solution for cos x = 0 will be x = (2n+1)π/2, n∈I, as cos x has a value equal to 0 at π/2, 3π/2, 5π/2, -7π/2, -11π/2 etc. Below here is the table defining the general solutions of the given trigonometric functions involved in equations.

Equations	Solutions
sin x = 0	x = nπ
cos x = 0	x = (nπ + π/2)
tan x = 0	x = nπ
sin x = 1	x = (2nπ + π/2) = (4n+1)π/2
cos x = 1	x = 2nπ
sin x = sin θ	x = nπ + (-1)ⁿθ, where θ ∈ [-π/2, π/2]
cos x = cos θ	x = 2nπ ± θ, where θ ∈ (0, π]
tan x = tan θ	x = nπ + θ, where θ ∈ (-π/2 , π/2]
sin 2x = sin 2θ	x = nπ ± θ
cos 2x = cos 2θ	x = nπ ± θ
tan 2x = tan 2θ	x = nπ ± θ

Proof of Solutions of Trigonometric Equations

Now let us prove these solutions here with the help of theorems.

Theorem 1: For any real numbers x and y, sin x = sin y implies x = nπ + (–1)n y, where n ∈ Z

Proof: Consider the equation, sin x = sin y. Let us try to find the general solution for this trigonometric equation.

sin x = sin y

⇒ sin x – sin y = 0

⇒2cos (x + y)/2 sin (x – y)/2 = 0

⇒cos (x + y)/2 = 0 or sin (x – y)/2 = 0

Upon taking the common solution from both the conditions, we get:

x = nπ + (-1)ⁿy, where n ∈ Z

Theorem 2: For any real numbers x and y, cos x = cos y, implies x = 2nπ ± y, where n ∈ Z.

Proof: Similarly, the general solution of cos x = cos y will be:

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 = nπ

On taking the common solution from both the conditions, we get:

x = 2nπ± y, where n ∈ Z

Theorem 3: Prove that if x and y are not odd multiple of π/2, then tan x = tan y implies x = nπ + y, where n ∈ Z.

Proof: Similarly, to find the solution of equations involving tan x or other functions, we can use the conversion of trigonometric equations.

In other words, if tan x = tan y then;

sinx/cosx = siny/cosy

sin x cos y = sin y cos x

sin x cos y – sin y cos x = 0

sin (x – y) = 0 [By trigonometric identity]

Hence, x – y =nπ or x = nπ + y, where n ∈ Z.

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 = 0

Taking 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 = nπ

Taking 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) = 0

As 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)ⁿy, 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.

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/3

As 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.

Trigonometric Equations Solved Examples

Let us go through an example to have a better insight into the solutions of trigonometric equations.

Q.1 : 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 = ½

⇒4x = nπ or 2x = 2nπ ± π/3

Therefore, the general solution for the given trigonometric equation is:

⇒x = nπ/4 or nπ ± π/6

Q.2: Find the principal solution of the equation sin x = 1/2.

Solution: Since we know, sin π/6 = 1/2

and sin 5π/6 = sin (π – π/6) = sin π/6 = 1/2

Therefore, the principal solutions are x =π/6 and x = 5π/6.

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.

MCQs on Trigonometric Equations – Class 11 Maths

These multiple-choice questions (MCQs) are based on NCERT Class 11 Maths Chapter: Trigonometric Equations and are useful for CBSE Board Exams and JEE preparation.

1. The general solution of sin⁡x=0 is:

A) x=nπ, n∈Z
B) x=2nπ, n∈Z
C) x=(2n+1)π, n∈Z
D) x=nπ/2, n∈Z

✅ Answer: A) x=nπ, n∈Z

2. The general solution of cos⁡x=0 is:

A) x=nπ, n∈Z
B) x=2nπ, n∈Z
C) x=(2n+1)π/2, n∈Z
D) x=nπ/2, n∈Z

✅ Answer: C) x=(2n+1)π/2, n∈Z

3. The general solution of tan⁡x=0 is:

A) x=nπ, n∈Z
B) x=2nπ, n∈Z
C) x=(2n+1)π, n∈Z
D) x=nπ/2, n∈Z

✅ Answer: A) x=nπ, n∈Z

4. The smallest positive solution of sin⁡x=1/2 is:

A) π/3
B) π/4
C) π/6
D) π/2

✅ Answer: C) π/6

5. The general solution of sin⁡x=sin⁡a is:

A) x=2nπ+ax, n∈Z
B) x=nπ+(−1)ⁿ, n∈Z
C) x=2nπ−ax , n∈Z
D) x=nπ−(−1)ⁿ , n∈Z

✅ Answer: B) x=nπ+(−1)ⁿ , n∈Z

6. The general solution of tan⁡x=tan⁡a is:

A) x=nπ+ax , n∈Z
B) x=2nπ+ax , n∈Z
C) x=nπ−ax , n∈Z
D) x=(2n+1)π+ax , n∈Z

✅ Answer: A) x=nπ+ax , n∈Z

7. The number of solutions of cos⁡x=1 in the interval [0,2π] is:

A) 0
B) 1
C) 2
D) Infinite

✅ Answer: B) 1 (Since cos⁡x=1 at x=0 in [0,2π]

8. The general solution of cos⁡2x=cos⁡x is:

A) x=2nπ
B) x=2nπ±π/3
C) x=2nπ±π/4
D) x=2nπ±π/2

✅ Answer: D) x=2nπ±π/2

9. The smallest positive root of tan⁡x=1 is:

A) π/6
B) π/4
C) π/3
D) π/2

✅ Answer: B) π/4 (Since tan⁡45^∘=1)

10. The number of solutions of sin⁡x=1/2 in the interval [0,2π] is:

A) 1
B) 2
C) 3
D) 4

✅ Answer: B) 2 (Solutions: x=π/6, 5π/6 in [0,2π])

Assertion & Reason Questions on Trigonometric Equations – Class 11 Maths

These Assertion and Reason questions are designed for CBSE Board Exams, JEE, and competitive exams based on NCERT Class 11 Maths Chapter: Trigonometric Equations.

1. Assertion (A): The general solution of sin⁡x = 0 is x = nπ, n∈Z.

Reason (R): The sine function is zero at integral multiples of π.