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]  

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

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)

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Nov 23, 24 10:01 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 76)

    Nov 23, 24 09:45 AM

    digitalsatmath63.png
    Digital SAT Math Problems and Solutions (Part - 76)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 75)

    Nov 21, 24 06:13 AM

    digitalsatmath62.png
    Digital SAT Math Problems and Solutions (Part - 75)

    Read More