Find the solutions to sin^2 (x) + cos(x) = 1, keeping 0 ≤ x < 2π

Question

A)x = 0, π/2, 3π/2 B)x = π/2, 3π/2, 2π C)x = 0, −3π/2, 3π/2 D)x = π/6, π/4, π/3

It would be much appreciated if someone could explain how to get the answer.

Answer 1

Recall that

 

sin²x = 1 - cos²x

 

(1 - cos²x) + cos(x) = 1

 

-cos²x + cos(x) = 0

 

Let y = cos(x)

 

-y² + y = 0

 

y(1 - y) = 0

 

y = 1 or y = 0

 

y = 1 ⇒ cos(x) = 1 ⇒ x = 0

 

y = 0 ⇒ cos(x) = 0 ⇒ x = π/2 or 3π/2