SOLVING TRIGNOMETRIC EQUATIONS ON THE INTERERVAL 0 AND 2PI
The equation containing trigonometric functions of unknown angles are known as trigonometric equations. A solution of a trigonometric equation is the value of unknown angle that satisfies the equation.
General solution :
The solution of a trigonometric equation giving all the admissible values obtained with the help of periodicity of a trigonometric function is called the general solution of the equation.
Principal Solution :
The smallest numerical value of unknown angle satisfying the equation in the interval [0, 2π] (or) [−π, π] is called a principal solution. We shall take the interval [−π, π] for defining the principal solution.
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Trigonometric equation sin θ = 0 cos θ = 0 tan θ = 0 |
General solution θ = nπ, n ∈ z θ = (2n + 1)π/2, n ∈ z θ = nπ, n ∈ z |
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sin θ = sin α α ∈ [-π/2, π/2] |
θ = nπ + (-1)n α, n ∈ z |
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cos θ = cos α α ∈ [0, π] |
θ = 2nπ ± α, n ∈ z |
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tan θ = tan α α ∈ (-π/2, π/2) |
θ = nπ + α, n ∈ z |
Solve for x ∈ [0, 2π] giving your answer as exact values.
Problem 1 :
2 sin2x + sin x = 0
Solution :
2 sin2x + sin x = 0
Let t = sin x
2t2 + t = 0
t(2t + 1) = 0
t = 0 and t = -1/2
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t = 0 sin x = 0 x = sin-1(0) x = nπ If n = 0, 1, 2 x = 0, π, 2π |
sin x = -1/2 x = sin-1(-1/2) x = -π/6 ∈ [-π/2, π/2] x = nπ + (-1)n α When n = 1 x = π + (-1)1 (-π/6) x = 7π/6 n = 2 x = 2π + (-1)2 (-π/6) x = 11π/6 |
So, the solutions are x = 0, π, 7π/6, 11π/6, 2π
Problem 2 :
2 cos2x = cos x
Solution :
2 cos2x - cos x = 0
cos x (2cos x - 1) = 0
cos x = 0 and 2cos x - 1 = 0
When cos x = 0
x = cos-1(0)
x = (2n + 1)π/2
n = 0, 1, ......
x = π/2, 3π/2
When 2cos x - 1 = 0
x = cos-1(1/2)
principal value
α = π/3
To get multiple angles
x = 2nπ ± α
n = 0, 1, 2, .......
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n = 0 x = ± π/3 |
n = 1 x = 2π± π/3 x = 7π/3 and x = 5π/3 Here 7π/3 is not in the interval. |
x = π/3, π/2, 3π/2, 5π/3
Problem 3 :
2 cos2x + cos x - 1 = 0
Solution :
2 cos2x + cos x - 1 = 0
Let t = cos x
2t2 + t - 1 = 0
(2t - 1) (t + 1) = 0
t = 1/2 and t = -1
When cos x = 1/2
Principal value α = π/3
x = 2nπ ± α
n = 0, 1, 2, ..........
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When n = 0 x = 2(0)π ± π/3 x = π/3 |
When n = 1 x = 2(1)π ± π/3 x = 7π/3, 5π/3 |
7π/3 does not belongs to the interval o to 2π.
When cos x = -1
x = π
So, the values of x are π/3, π, π/3
Problem 4 :
2 sin2x + 3sinx + 1 = 0
Solution :
2 sin2x + 3sinx + 1 = 0
Let t = sin x
2t2 + 3t + 1 = 0
(2t + 1)(t + 1) = 0
t = -1/2 and t = -1
When sin x = -1/2
x = nπ + (-1)n α
Principal value α = -π/6
n = 0, 1, 2, ........
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n = 0 x = -π/6 |
n = 1 x = π + (-1)1 (-π/6) x = 7π/6 |
n = 2 x = 2π + (-1)2 (-π/6) x = 11π/6 |
When sin x = -1
x = 3π/2
So, the values are x = 7π/6, 3π/2, 11π/6
Problem 5 :
sin2x = 2 - cos x
Solution :
sin2x = 2 - cos x
1 - cos2x = 2 - cos x
cos2x - cos x + 2 - 1 = 0
cos2x - cos x + 1 = 0
Let t = cos x
t2 - t + 1 = 0
a = 1, b = -1 and c = 1
There is no real value. So, there is no solution.
Problem 6 :
3 tan x = cot x
Solution :
3 tan x = cot x
3 tan x = 1/tan x
3tan2x = 1
tan x = 1/√3
tan x = ±1/√3
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tan x = 1/√3 θ = nπ + α Principal value α = π/6 When n = 0, 1, 2......... n = 0 x = π/6 n = 1 x = 7π/6 |
tan x = -1/√3 θ = nπ + α Principal value α = -π/6 When n = 0, 1, 2......... n = 0 x = -π/6 n = 1 x = 5π/6 n = 2 x = 11π/6 |
So, the values of x are π/6, 5π/6, 7π/6, 11π/6
Problem 7 :
sin 4x = sin 2x
Solution :
sin 4x = sin 2x
sin2(2x) = sin 2x
2sin2x cos 2x - sin 2x = 0
sin 2x(2 cos 2x - 1) = 0
sin 2x = 0 and 2 cos 2x - 1 = 0
When sin 2x = 0, x = sin-1(0)
x = 0, π/2, 3π/2, 2π
2 cos2x = 1
cos 2x = 1/2
1-2sin2x = 1/2
2sin2x = 1 - (1/2)
2sin2x = (1/2)
sin2x = (1/4)
sin x = √(1/4)
sin x = ±1/2
sin x = 1/2 and sin x = -1/2
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sin x = 1/2 x = sin-1(1/2) x = nπ + (-1)n α Principal value α = π/6 n = 0, 1, 2, ...... n = 0 x = π/6 n = 1 x = π + (-1)1 (π/6) x = 5π/6 |
sin x = -1/2 x = sin-1(-1/2) x = nπ + (-1)n α Principal value α = -π/6 n = 0, 1, 2, ...... n = 0 x = -π/6 n = 1 x = π + (-1)1 (-π/6) x = 7π/6 n = 1 x = 2π + (-1)2 (-π/6) x = 11π/6 |
So, the solutions are
x = 0, π/6, π/2, 5π/6, π, 7π/6 3π/2, 11π/6, 2π
Problem 8 :
sin x + cos x = √2
Solution :
sin x + cos x = √2
Take square on both sides.
(sin x + cos x)2 = √22
sin2x + cos2x + 2sin x cos x = 2
1 - cos2x + cos2x + sin 2x = 2
sin 2x = 1
2x = sin -1(1)
2x = π/2, 3π/2
x = π/4 and 3π/4
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