Principal solution for sin x = ½

  sin x = ½

Here sin is positive,

Principal solution for sin x .jpg

We know that

  sin is positive in 1st and 2nd quadrant

 

Value in 1st Quadrant = 30°

Value in 2nd Quadrant = 180° – 30° = 150°

 

So, Principal solutions are

  x = 30° = 30° × π/180 = π/6

  x = 150° = 150° × π/180 = 5π/6

 

Thus, Principal solutions are

  π/6 & 5π/6

 

Principal solution for cos x = –1/√2

  cos x = –1/√2

Here cos is negative,

Principal solution for sin x .jpg

We know that

  cos is negative in 2nd and 3rd quadrant

Here, θ = 45°

 

Value in 2nd Quadrant = 180° – 45° = 135°

Value in 3rd Quadrant = 180° + 45° = 225°

 

So, Principal solutions are

  x = 135° = 135° × π/180 = 3π/4

  x = 225° = 225° × π/180 = 5π/4

 

Thus, Principal solutions are

  3π/4 & 5π/4

 

Principal solution for tan x = –1

  tan x = –1

Here tan is negative,

Principal solution for sin x .jpg

We know that

  tan is negative in 2nd and 4th quadrant

Here, θ = 45°

 

Value in 2nd Quadrant = 180° – 45° = 135°

Value in 4th Quadrant = 360° – 45° = 315°

 

So, Principal solutions are

  x = 135° = 135° × π/180 = 3π/4

  x = 315° = 315° × π/180 = 7π/4

 

Thus, Principal solutions are

  3π/4 & 7π/4

  1. Chapter 3 Class 11 Trigonometric Functions
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Transcript

Principal solution for sin x = ½   sin x = ½ Here sin is positive, We know that   sin is positive in 1st and 2nd quadrant Value in 1st Quadrant = 30° Value in 2nd Quadrant = 180° – 30° = 150° So, Principal solutions are   x = 30° = 30° × π/180 = π/6   x = 150° = 150° × π/180 = 5π/6 Thus, Principal solutions are   π/6 & 5π/6 Principal solution for cos x = –1/√2   cos x = –1/√2 Here cos is negative, We know that   cos is negative in 2nd and 3rd quadrant Here, θ = 45° Value in 2nd Quadrant = 180° – 45° = 135° Value in 3rd Quadrant = 180° + 45° = 225° So, Principal solutions are   x = 135° = 135° × π/180 = 3π/4   x = 225° = 225° × π/180 = 5π/4 Thus, Principal solutions are   3π/4 & 5π/4 Principal solution for tan x = –1   tan x = –1 Here tan is negative, We know that   tan is negative in 2nd and 4th quadrant Here, θ = 45° Value in 2nd Quadrant = 180° – 45° = 135° Value in 4th Quadrant = 360° – 45° = 315° So, Principal solutions are   x = 135° = 135° × π/180 = 3π/4   x = 315° = 315° × π/180 = 7π/4 Thus, Principal solutions are   3π/4 & 7π/4

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Davneet Singh
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