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.
Principal value of sine function lies in the interval
[−π/2, π/2]
and hence lies in I quadrant or IV quadrant.
Principal value of cosine function is in
[0, π]
and hence in I quadrant or II quadrant.
Principal value of tangent function is in
(-π/2, π/2)
and hence in I quadrant or IV quadrant.
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.
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 |
Question 1 :
Find the principal solution and general solutions of the following :
sin θ = -1/√2
Solution :
sin θ = -1/√2
θ = sin^{-1}(-1/√2)
Principal Solution :
We have to choose the principal solution between the interval [-π/2, π/2] .
θ = -π/4
General Solution :
θ = nπ + (−1)^{n} α, n ∈ Z
θ = nπ + (−1)^{n}(-π/4), n ∈ Z
Question 2 :
Find the principal solution and general solutions of the following :
cot θ = √3
Solution :
cot θ = √3
1/tan θ = 1/√3
Taking reciprocals on both sides
tan θ = √3
Principal Solution :
We have to choose the principal solution between the interval (−π/2, π/2).
θ = π/6
General Solution :
θ = nπ + α, n ∈ Z
θ = nπ + (π/6)
Question 3 :
Find the principal solution and general solutions of the following :
tan θ = -1/√3
Solution :
tan θ = -1/√3
Principal Solution :
We have to choose the principal solution between the interval (-π/2, π/2).
θ = -π/6
General Solution :
θ = nπ + α, n ∈ Z
θ = nπ + (-π/6)
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