Finding principal solutions

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