# PRINCIPAL SOLUTION AND GENERAL SOLUTION OF TRIGONOMETRIC FUNCTIONS

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]

Principal value of cosine function is in

[0, π]

Principal value of tangent function is in

(-π/2, π/2)

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 equationsin θ = 0cos θ = 0tan θ = 0sin 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 solutionθ = nπ; n ∈ Zθ = (2n + 1) π/2; n ∈ Zθ = nπ; n ∈ Zθ = nπ + (−1)n α, n ∈ Zθ = 2nπ ± α, n ∈ Zθ = nπ + α, n ∈ Z

## Practice Questions

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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