FIND THE EXACT SOLUTIONS OF THE GIVEN TRIGONOMETRIC EQUATION
The equations containing trigonometric functions of unknown angles are known as trigonometric equations. A solution of 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.
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Trigonometric equation sin θ = 0 cos θ = 0 tan θ = 0 sin θ = sinα, where α ∈ [−π/2, π/2] cos θ = cos α, where α ∈ [0,π] tan θ = tanα, where α ∈ (−π/2, π/2) |
General solution θ = nπ; n ∈ Z θ = (2n + 1) π/2; n ∈ Z θ = nπ; n ∈ Z θ = nπ + (−1)n α, n ∈ Z θ = 2nπ ± α, n ∈ Z θ = nπ + α, n ∈ Z |
Practice Questions
Question 1 :
Solve the following equation :
cos θ + cos 3θ = 2cos 2θ
Solution :
cos θ + cos3θ = 2cos2θ
Let us use the formula for cos C + cos D
cos C + cos D = 2 cos (C + D)/2 cos (C - D)/2
cos θ + cos3θ = 2cos2θ
2 cos 2θ cos θ - 2 cos 2θ = 0
2 cos 2θ (cos θ - 1) = 0
2 cos 2θ = 0 cos θ - 1 = 0
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2 cos 2θ = 0 cos 2θ = 0 2θ = cos-1(0) 2θ = (2n + 1)π/2 θ = (2n + 1)π/4 |
cos θ - 1 = 0 cos θ = 1 α = 0 θ = 2nπ ± α, n ∈ Z θ = 2nπ ± 0, n ∈ Z θ = 2nπ , n ∈ Z |
So, the solution is {(2n + 1)π/4, 2nπ}.
Question 2 :
Solve the following equation :
sin θ + sin 3θ + sin 5θ = 0
Solution :
sin θ + sin 3θ + sin 5θ = 0
First let us use the formula for sin C + sin D
2 sin 3θ cos 2θ + sin 3θ = 0
sin 3θ (2cos 2θ + 1) = 0
sin 3θ = 0 (or) 2cos 2θ + 1 = 0
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sin 3θ = 0 3θ = nπ; n ∈ Z θ = nπ/3; n ∈ Z |
2cos 2θ + 1 = 0 2cos 2θ = -1 cos 2θ = -1/2 a = π - (π/3) ==> 2π/3 2θ = 2nπ ± α, n ∈ Z 2θ = 2nπ ± 2π/3, n ∈ Z θ = nπ ± π/3, n ∈ Z |
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