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Trigonometric Equation and its Solutions 

Trigonometric Equation 

An equation involving one or more trigonometric ratios of unknown angles is called a trigonometric equation. A trigonometric equation can be written as Q1 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ) = Q2 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ), where Q1 and Q2 are rational functions. 

Example: Let us consider an equation cos2x – 4 sin x = 1.

This is a trigonometric equation and not an identity as it is not satisfied for all values of x e.g. the equation is not satisfied at (2n + 1)π/4. 

Solution of a trigonometric equation:

All possible values of unknown which satisfy the given equation are called solution of the given equation.

For complete solution “all possible values” satisfying the equation must be obtained. 

When we try to solve a trigonometric equation, we try to find out all sets of values of θ, which satisfy the given equation. Sometimes, in simple equations and when it is easy to draw a graph of an equation, one can find out the solution simply by viewing the graph. 

 

Period of a function:

A function f(x) is said to be periodic if there exists T > 0 such that f (x+T) = f(x) for all x in the domain of definition of f(x). If T is the smallest positive real number such that f (x+T) = f(x), then it is called the period of f(x).

The trigonometric functions such as sin, cos and tan are periodic functions. 

Illustration: We try to search for the solutions of the equation sin θ = 0 other than θ = 0. By seeing the equation, one might straight away reach at the conclusion that θ = 0 is the only solution. But, in case of trigonometric equations, it is important to rule out all possibilities so as to reach at the correct solution.

Let OX be the initial line

Let ∠POX = θ and OP = r                                          

From ΔPOL,

sin θ = PL/OP = y/r.

Now sin θ = 0

⇒ y/r = 0; ⇒ y = 0.

This is possible only when OP coincides with OX or OX’.

When OP coincides with OX, θ = 0, ± 2π, ± 4π and ± 6π ……… (1)

And when OP coincides with OX’, θ = ± π, ± 3π, ± 5π ……… (2)

Thus from (1) and (2) it follows that at sin θ = 0                               

θ = nπ, where n = 0, ±1, ±2, ………

We call θ = nπ, a general solution of the trigonometric equation sin θ = 0, because for all values of n, this solution satisfies the given equation.

Illustration: General solution of cos θ = 0 

cos θ = 0 ⇒ x = π/2.

This is possible only when OP coincides with OY or OY’

When OP coincides with OY,

θ = π/2, 5π/2, 9π/2 or, -3π/2, -7π/2  ..……… (1) 

when OP coincides with OY’

θ = -3π/2, -7π/2 or, -π/2, -5π/2       ………… (2)                                                   

Thus from (1) and (2) if follows that the general solution of cos θ = 0 is θ (2n+1) π/2, where n = 0, ±1, ±2 ……… 

For more on trigonometric equations, please refer the video given below:

General solution of the equation sin θ = k. 

We know that when sin θ = k, k has to be such that –1 ≤ k ≤ 1 

We can always find some α ∈ [–π/2, π/2]

As sin (-π)/2 = -1 & sin π/2 = 1, such that sin θ = k, i.e. α = sin-1k

i.e. sin θ = sin α, α ∈ [–π/2, π/2]

⇒ sin θ – sin α = 0

⇒ 2 sin {(θ – α)/2} cos {θ + α)/2} = 0 

from the above equation to be satisfied, either sin {(θ – α)/2) = 0

and consequently ((θ – α)/2) = integral multiple of π

∴ θ – α = 2nπ

i.e. θ = 2nπ + α

θ = 2nπ + (–1)2n α where n = 0, ±1, ±2  … (1) 

or, cos {(θ + α)/2} = 0

i.e. {(θ + α)/2} = any odd multiple of π/2

i.e. {(θ + α)/2} = (2n + 1)π/2 

i.e. θ = (2n + 1)π – α 

⇒ θ = (2n +1)π + (–1)2n+1 α … (2)

From (1) and (2), we conclude that

θ = nπ + (–1)n α, where n is integral multiple, is the general solution of the equation sin θ = k 

Trigonometric Equations with their general Solutions:

Trigonometrical equation

General Solution

sin θ = 0 

 Then 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]

sin2 θ = sin2 α

 θ = nπ ± α

cos2 θ = cos2 α

 θ = nπ ± α

tan2 θ = tan2 α

 θ = nπ ± α

If α is assumed to be the least positive value of θ which satisfies two given trigonometrical equations, then the general value of θ will be 2nπ + α.  

Illustration: Find the general solution of equation sin θ = 1/2 

Solution: We know that sin θ = 1/2 = sin π/6.

So, general solution of the given equation is θ = nπ + (–1)nπ/6, n ∈ 0, ±1, ±2 

Illustration: Solve the equation sin 6x + sin 4x = 0. 

Solution: Applying the formulae for the sum of sine, i.e.

sin A + sin B = sin (A+B)/2 . cos (A-B)/2, we have

sin 5x cos x = 0             ……… (1) 

If ‘x’ is a solution for the equation, then at least one of the following equations is true:

sin 5x = 0 or cos x = 0  ……… (2)

Conversely, if x is a solution of one of the equation (2) then it is solution of equation (1) as well. Thus equation (1) is equivalent to the equation (2). The solutions of the equation (2) are given by

x = nπ/5, x = (2n+1)π/2, where n = 0, ±1, ±2 ……

All these values of x and only these values are the solutions of the original equation.

Illustration: Find the most general value of θ where sin θ = – √3/2 and tan θ = √3. 

Solution: Signs assume great importance in case of trigonometric functions. Students generally tend to mention the general solutions of sin and tan θ, which is wrong as that would not give us the entire solution.

sin θ is negative in 3rd and 4th Quadrant and tan θ is positive in 1st and 3rd Quadrant.

So common is 3rd Quadrant and at θ = 4π/3 both are satisfied.

∴ The general solution is 2nπ + 4π/3. 

This is because in interval [0, 2π] it is satisfied only at 4π/3. Again in [2π, 4π] it is satisfied at 2π + 4π/3 and so on. 

Hence, the general solution of the equation is 2nπ + 4π/3.

Illustration: Find general solution of cos 3θ = sin 2θ. 

Solution: This can be solved by 2 different ways.

Method 1: We can write the given equation as 

cos 3θ = cos (π/2 - 2θ)

⇒ 3θ = 2nπ + (π/2 - 2θ), where n = 0, ±1, ±2 …… 

or 5θ = 2nπ + π/2 and also θ = 2nπ – π/2

or θ = (4n + 1) π/10 and           

θ = (4n–1) π/2, where n ∈ I …… (A) 

Method 2: sin 2θ = sin (π/2 - 3θ)

2θ = nπ + (–1)n (π/2 - 3θ). 

Case I: When n is even, n = 2m, where m = 0, ±1, ±2 ……

2θ = 2mπ + π/2 - 3θ

θ = (4m+1)π/10, where m ∈ I ……. (B) 

Case II: When n is odd, n = (2m + 1)

2θ = (2m+1)π – (π/2 - 3θ)

θ = – (4m + 1) π/2, where m = 0, ±1, ±2 …… (B) 

Note: No doubt solutions obtained by both methods for odd values of n are different but as shown in the chart below, you can see that all possible values of θ are obtainable by both the given solutions:

             From B

              From A

  for m = 0, θ = – π/2,  

  for n = 0, θ = – π/2 

  for m = 1, θ = – 5π/2,           

  for n = 1, θ = + 3π/2 

  for m = 2, θ = – 9π/2,           

  for n = 2, θ = + 7π/2 

  for m = –1, θ = – 3π/2,         

  for m = –1, θ = – 5π/2 

  for m = –2, θ = 7π/2

  for m = –2, θ = – 9π/2 

General solution of sin2 θ = k, where k ∈ [0, 1] 

Given that sin2 θ = k, k ∈ [0, 1] 

We can find some α such that 

⇒ sin2 θ = sin2α where α = sin-1 √k

i.e. (sin θ – sin α) (sin θ + sin α) = 0

either sin θ – sin α = 0

θ = nπ + (–1)nα, where n = 0, ±1, ±2 ……… (1) 

or, sin θ + sin α = 0

sin θ = – sin α 

θ = nπ – (–1)nα, where n = 0, ±1, ±2     ……..(2) 

From (1) and (2) we get the general solution of equation the given 

θ = nπ ± α where n = 0, ±1, ±2 …… and α = sin-1 √k

Illustration: Solve the equation 7tan2 θ – 9 = 3 sec2θ

Solution: Given, 7tan2θ – 9 = 3 sec2θ

or, 7tan2θ – 9 = 3(1 + tan2θ) 
or, 4tan2θ = 12 
or, tan2θ = 3 
or, tan2θ = (tan π/3)2 
⇒ θ = nπ + π/3, where n = 0, ±1, ±2 ………… 

Note: We cannot define a unique method of solving trigonometric equations. In each case, success in solving a trigonometric equation depends, in particular, on the knowledge and application ability of trigonometric formula and the practice of solving problems. Many trigonometric formulas are true equalities for all the values of the variable’s appearing in them. 

Illustration: Solve the equation: cos θ = 0 
Solution: We can solve it to get two forms 
cos θ = 0 ⇒ θ = (2n + 1)π/2

cos θ = cos π/2 ⇒ θ = 2nπ + π/2 
or, θ = (4n + 1)π/2. 

Important: Following tips and steps will help you systematically solve trigonometric equations.

1. Try to reduce equation in terms of one single trigonometric ratios preferably sin θ or cos θ. 
If we have choice to convert a problem in sine or cosine then  cosine form is convenient compared to sine form. This is because in general solution of sine, we will have to deal with (–1)n which is  inconvenient compared to the dealing of + obtained in cosine form.
2. Factorize the polynomial in terms of these ratios. 
3. For LHS to be zero solve for each factor. And write down general solution, for each factor based on the standard results that are derived earlier in this section.

e.g. sin θ – k1 = 0 ⇒ θ = nπ + (–1)n sin-1 k1

cos θ – k2 = 0 ⇒ θ = 2nπ + cos–1 k2

Caution: You must check that k1, k2 ∈ [–1, 1]. Don’t blindly write it as it is since it will be absurd if they do not belong to [–1, 1].

Illustration: Solve the equation 5sin θ – 2 cos2θ – 1 = 0

Solution: Given, 5 sin θ – 2 cos2θ – 1 = 0

or, 5 sin θ – 2 (1 – sin2θ) – 1 = 0 
or, 2 sin2 θ + 5 sin θ – 3 = 0 
or, (sin θ + 3) (2 sin θ – 1) = 0 
∴ sin θ = -3 or sin θ = ½

First consider the case when sin θ = -3.

But this case is not possible as the range of sine is [-1, 1].

When sin θ = ½

Then we have sin θ = sin π/6.

⇒ θ = nπ + (–1)n π/6 where n = 0, ±1, ±2 ……… 

Note: Never divide by any expression which is zero. e.g. If the given equation is (sin θ – cos θ) (A) = (B) where A and B denote trigonometric equations then you can divide by (sin θ – cos θ) only when θ ≠ nπ + π/4

Illustration: Solve the equation tan θ + sec θ = √3 

Solution: tan θ + sec θ = √3                   …… (1)

Then (sin θ)/(cos θ) +1/(cos θ) = √3       …… (2)

or, cos (θ + π/6) = cos π/3

General solution θ + π/6 = 2nπ ± π/3, n ∈ I 

Taking positive sign, θ + π/6 = 2nπ + π/3

⇒ θ = 2nπ + π/6

Taking negative sign, θ + π/6 = 2nπ – π/3

⇒ θ = 2nπ – π/2

i.e. θ = (4n – 1)π/2.

But the solution obtained is correct only if, cos θ ≠ 0 otherwise (2) is not defined.

i.e. θ ≠ odd multiple of π/2

⇒ θ ≠ (4n – 1) π/2.

Hence the general solution will be θ = 2nπ + π/6 only where n = 0, ±1, ±2 …… 

Note: Domain of an equation should not change. In case it changes, necessary changes must be made in the general solution. 

Illustration: Solve the equation tan 5θ = tan 3θ 

Solution: Now tan 5θ = tan 3θ 

⇒ 5θ = nπ + 3θ 

or, 2θ = nπ

θ = nπ/2, where n = 0, ±1, ±2 …… 

putting n = 0 gives θ = 0, original equation is satisfied 

putting n = 1 we obtain θ = π/2 equation becomes tan 5π/2 = tan 3π/2.

Equation is not defined for odd multiple of π/2.

Hence we conclude that θ = 2nπ, where n = 0, ±1, ±2 ……… 

Some key points to be noted:  

1. If tan θ or sec θ is involved in the equation, θ ≠ odd multiple of π/2. 

2. If cot θ or cosec θ is involved in the equation, θ ≠ multiple of π or 0.

Trigonometry is full of formulas and the students are advised to learn all the trigonometric formulas including the trigonometry basics so as to remain competitive in JEE and other engineering exams. Students must practice various trigonometry problems based on trigonometric ratios and trigonometry basics so as to get acquainted with the topic.      

You might like to refer some of the related resources listed below:

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