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Theorems helpful to solve trigonometric equations
1. sin x = sin y ⇒ x = nπ + (-1)ny, where x, y are real and n ∈ Z.
2. cos x = cos y ⇒ x = 2nπ ± y
3. tan x = tan y ⇒ x = nπ + y, where 'x' and 'y' are not odd multiples of .
Example
Solve: sin x = - sin2 x.
Sol:
sin x = - sin2 x
sin x + sin2 x = 0
sin x(1 + sin x) = 0
sin x = 0 or 1 + sin x = 0
• If sin x = 0, then x = nπ where n ∈ Z.
• If 1 + sin x = 0, then sin x = -1 and so x = nπ + (-1)n(-) where n ∈ Z.
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.

Example
Find the principal solution of the following equations.
i. sin θ = √3/2
ii. cos 3θ = –1/2
iii. tan 5θ = 1/√3
Sol :
i. We have sin θ = √3/2
sin θ = sin π/3
θ = π/3 is the principal solution
∵ π/3 ∈ [– , ]
ii. We have cos 3θ = – 1/2
cos 3θ = cos 2π/3
3θ = 2π/3
2π/3 is the principal solution of 2θ
∵ 2π/3 ∈ [0, π]
iii. We have tan 5θ = 1/√3
tan 5θ = tan
5θ =
∴ is the principal solution of 5θ
∵ ∈ (– , ).
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}.


MODEL TESTS
EAMCET (ENGG) JEE (MAIN) JEE (ADV) NEET

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