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Trigonometric Equations

Equations involving one or more trigonometric functions of a variable are called trigonometric equations. As you know by now, the values of 'sin x' and 'cos x' repeat after an interval of 2π rad (or 360°) whereas the values of 'tan x' repeat after an interval of π (or 180°).

If the trigonometric equations are satisfied by all values of unknown angles, for which the functions are defined, then they are called identities.

The values of unknown angle that satisfy the equation are called solutions. The set of all solutions is called the solution set or general solution of a trigonometric equation. In other words, a general solution is an expression involving integer 'n' which gives all the solutions.

The solution in which the absolute value of the angle is the least is called principal solution. In other words, the solutions of a trigonometric equation for which 'x' lies in the interval 0 to 2π, i.e., 0 ≤ x ≤ 2π, are called the principal solution.

The process of finding the solution set is called solving the trigonometric equation.

We know that trigonometric functions are periodic function. Hence the trigonometric equations may have infinite number of solutions.

It is not necessary that every trigonometric equation has a solution.
For example: sin θ = 4 has no solution.

For any integer n, 2nπ + θ is also a solution of given trigonometric equation.

Principal solution of an equation

Principal solution of an equation (or Principal value of an angle)

i. The sine function from [–, ] → [– 1, 1] is bijective.
There exists a unique value 'α' in [–, ] satisfying sin α = k for k ∈ [– 1, 1].
This 'α' is called principal value or principal solution of 'θ' satisfying the equation sin θ = k.

ii. The cosine function from [0, π] → [– 1, 1] is bijective.
There exists a unique value 'α' in [0, π] satisfying cos α = k for k ∈ [– 1, 1].
This 'α' is called the principal solution of 'θ' satisfying the equation cos θ = k.

iii. The tangent function from (–, ) → (– ∞, ∞) is bijective.
There exists a unique value 'α' in (–, ) satisfying tan α = k for k ∈ (– ∞, ∞).
This 'α' is called the principal solution of 'θ' satisfying the equation tan θ = k.

iv. The principal value of θ satisfying the equation
a) cot θ = k, k ∈ (– ∞, ∞), lies in [–, ] – {0}
b) sec θ = k, k ∈ ( – ∞, – 1] ∪ [1, ∞) lies in [0, π] – {}
c) cosec θ = k, k ∈ ( – ∞, – 1] ∪ [1, ∞) lies in [–, ] – {0}.

General Solutions

General solutions of the trigonometric equations

Type-I

Pre-requisite: Domain, range and graphs of trigonometric functions discussed earlier

i. The general solution of sin θ = 0 is given by
θ = nπ, n ∈ z

ii. The general solution of cos θ = 0 is given by
θ = (2n + 1), n ∈ z

iii. The general solution of tan θ = 0 is given by
θ = nπ, n ∈ z

iv. The general solution of cot θ = 0 is given by
θ = (2n + 1), n ∈ z

Type-II

i. General solution of the equation sin θ = k, k ∈ [– 1, 1] is
θ = nπ + (–1)n α
where n ∈ z, 'α' is the principal solution, α ∈ [–, ].

ii. General solution of the equation cos θ = k, k ∈ [– 1, 1] is
θ = 2nπ ± α
where n ∈ z and α lies in [0, π].

iii. General solution of tan θ = k is
θ = nπ + α
where k ∈ R, n ∈ z and α lies in (–, )

For general solutions of reciprocal functions of the above three ratios refer adjacent table.

Type-III

General solutions of trigonometric equations of the form
sin2 θ = sin2 α, cos2 θ = cos2 α and tan2 θ = tan2 α

The difference is in the principal solution as under:

(i) sin2 θ = sin2 α ⇒ θ = nπ ± α, n ∈ z, where α is the principal solution of sin θ = sin α

(ii) cos2 θ = cos2 α ⇒ θ = nπ ± α, n ∈ z, where α is the principal solution of cos θ = cos α

(iii) tan2 θ = tan2 α ⇒ θ = nπ ± α, n ∈ z, where α is the principal solution of tan θ = tan α

Type-IV

General solution for two or more simultaneous equations in one variable

Algorithm:

i. First find the value of unknown angle θ lying between 0 and 2π and satisfying the two or more given equations separately.

ii. Select the value of an angle θ which satisfies both equations.

iii. General solution is given by θ = 2nπ + α, n ∈ z and α is the principal value of θ lying between 0 and 2π.

Type-V

Trigonometric equations of the form
a cos θ + b sin θ = c, where a, b, c ∈ R such that | c | ≤

To solve this type of equations, we first reduce them in the form cos θ = cos α (or) sin θ = sin α

Algorithm :

i. Write the given equation a cos θ + b sin θ = c

ii. Put a = r cos α and b = r sin α, where r = and tan α = b/a
⇒ α = tan- 1(b/a)

iii. By substituting a, b values in equation (i), the equation becomes r cos (θ – α) = c
⇒ cos(θ – α) = c/r = cos β (say)

iv. Solve the above equation by using the formula discussed earlier

Note :

If | c | >
⇒ a cos θ + b sin θ = c has no solution.

Points to remember while solving trigonometric equations click More info

Trigonometric Inequalities

A trigonometric inequality is an inequality in the form of
f(x) ≤ a   or
f(x) ≥ a  
where f(x) contains one or more trigonometric function and 'a' is a constant.
Solving the inequality f(x) means finding all the values of the variable 'x' such that it satisfying the trigonometric inequality. The set of all values of the variable 'x' is called solution set.
Solution sets of trigonometric inequalities are expressed in intervals.
It is explained through typical examples below.

Ex 1: f(x) is a single trigonometric function as in tan x ≥ – 1
We observe that the given inequality contains two different functions. On the LHS is a trigonometric function, while a constant is at the RHS. Plot them in single coordinate system.(Refer adjacent image)

The corresponding trigonometric equation is
tan x = – 1 ⇒ x = – ∈ (– , )
Tangent values are greater than -1 for angle greater than -.
The basic interval satisfying the inequality is – ≤ x <
It is also clear that the solution in this interval is repeated with period of π (since tangent function is periodic with π). Hence, general solution of the given inequality is
nπ – ≤ x < nπ + ; n ∈ Z.

Examples of inequality
Example: f(x) is a combination of trigonometric functions as in sin x < cos x.
Sol: Convert sin x < cos x in terms of inequality of a single trigonometric function.
i.e, cos x – sin x > 0
cos x (1/√2) – sin x (1/√2) > 0 (1/√2)
cos x . cos – sin x . sin > 0
cos (x + ) > 0
We know that the cos function is positive in the Ist and IV th quadrants.
The base interval is (– , )
i.e, – < x + <
The periodicity of the cos function is 2π.
Hence, we add 2nπ on either side of the base interval.
2nπ – < x + < 2nπ + ; n ∈ Z
2nπ – (3π/4) < x < 2nπ + ; n ∈ Z
Graphical method of inequality
Example: If 0 ≤ x ≤ 2π and |cos x| ≤ sin x, then prove that x ∈ [, 3π/4].
Sol: To find the solution set, plot the graphs of the above functions in single coordinate system.
From the graph, clearly the solution set is [, 3π/4]
Process of solving trigonometric inequalities

Let us consider different cases and defined the process to solve the inequality.

Case I: The trigonometry inequality is of the form: f(x) ≥ a or f(x) ≤ a, where f(x) is any single trigonometric function and 'a' is any real number.

We may follow these steps:

1. Consider the two sides of the inequality as two functions.

2. Plot the two graphs of the above functions on a single scale.

3. Identity the region that satisfies the inequality.

4. Find the basic interval of inequality.

5. Add 2nπ or nπ based on period of the function.

This gives the general solution of the inequality.


Case II: Inequality is of the form: a ≤ f(x) ≤ b; where f(x) is any trigonometric function and a, b are any two real numbers.

We may follow these steps:

1. Divide the given inequality into two separate inequalities.

2. Solve them separately.

3. Find the common solution set of both the inequalities which is the solution of given inequality.

Case III: The trigonometry inequality is of the form: g(x) ≥ 0 or g(x) ≤ 0; where g(x) contains two or more trigonometric functions and 'a' is any real number.

We may follow these steps:

1. Transform the given trigonometric inequality into single trigonometric function form using trigonometric identities.

2. Find the common period, the common period must be the least multiple of the periods of all trigonometric functions presented in the inequality. The complete solution set must, at least cover one whole common period.

3. Solve the trigonometric equation g(x) = 0

4. Solve the inequality g(x) ≤ 0 or g(x) ≥ 0 by the algebraic method.

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