Solving trigonometrical equations is nothing but finding all replacements for the variable θ that make the equations true.
A solution of a trigonometrical equation is the value of the unknown angle that satisfies the equation. A trigonometrical equation may have infinite number of solutions.
Principal Solution :
The solution in which the absolute value of the angle is the least is called principal solution.
General Solution :
The expression involving integer ‘n’ which gives all solutions of a trigonometrical equation is called the general solution.
Example 1 :
Solve 2 cos²θ + 3 sin θ = 0
Solution :
2 cos²θ + 3 sin θ = 0
First we have to convert the given function in one trigonometric ratio.
cos²θ = 1 - sin²θ
2 (1 - sin²θ) + 3 sin θ = 0
2 - 2 sin²θ + 3 sin θ = 0
Multiply the whole equation by (-)
2 sin²θ - 3 sin θ - 2 = 0
2 sin²θ - 4 sin θ + sin θ - 2 = 0
2 sin θ (sin θ - 2) + 1(sin θ - 2) = 0
(2 sin θ + 1) (sin θ - 2) = 0
2 sin θ + 1 = 0 Subtract 1 on both sides 2 sin θ + 1 - 1 = 0 - 1 2 sin θ = -1 Divide by 2 on both sides sin θ = -1/2 |
sin θ - 2 = 0 sin θ = 2 it is not possible |
sin θ = - sin (π/6)
sin θ = sin (-π/6) [satisfies the condition sin θ = sin α]
So, α = θ = -π/6
Other angles are
General angle for sin θ :
θ = nπ + (-1)ⁿ α [n ∈ z ]
θ = nπ + (-1)ⁿ (-π/6)
For example
if we put n = 1, we will get
θ = π + (-1)¹ (-π/6) ==> π -π/6 ==> 5π/6
Hence θ = -π/6, 5π/6, ..........
Example 2 :
Solve 2 tan θ − cot θ = − 1
Solution :
2 tan θ − cot θ = − 1
2 tan θ − (1/tan θ) = − 1
(2 tan²θ − 1)/tan θ = − 1
Multiply by tan θ on both sides
(2 tan²θ − 1) = − tan θ
2 tan²θ + tan θ − 1 = 0
Let x = tan θ
2x² + x - 1 = 0
2x² + 2x - x - 1 = 0 ==> 2x(x + 1) - 1(x + 1) = 0
(2x - 1) (x + 1) = 0
2 x - 1 = 0 2 tan θ - 1 = 0 Add 1 on both sides 2 tan θ = 1 Divide by 2 on both sides tan θ = 1/2 |
x + 1 = 0 Subtract 1 on both sides x + 1 - 1 = 0 - 1 x = -1 tan θ = -1 |
tan θ = -1
tan θ = -tan (π/4)
tan θ = tan (-π/4) [satisfies the condition tan θ = tan α]
So, α = θ = -π/4
Other angles are
General angle for tan θ :
θ = nπ + (-1)ⁿ α [n ∈ z ]
θ = nπ -(π/4)
tan θ = 1/2
θ = tan ⁻¹(1/2)
θ = nπ + (-1)ⁿ α [n ∈ z ]
θ = nπ + tan ⁻¹(1/2)
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