cosx=1-sinx find all solution in interval [0,2pi)

Question

please help me find the solutions to this

Answer 1

Hi Nicole O.,

cos(x) = 1 - sin(x)

sin(x) + cos(x) = 1;

sin²(x) + 2sin(x)cos(x) + cos²(x) = 1;

sin²(x) + cos²(x) + 2sin(x)cos(x) = 1

1 + 2sin(x)cos(x) = 1;

2sin(x)cos(x) = 0

sin(2x) = 0;

Sin(θ) is zero at, θ = 0 and, θ = π, between [0, 2π).

2x = 0 and 2x = π;

x = 0 and x = π/2 for 0 ≤ x < 2π;

I hope this helps, Joe.

Answer 2

cos x = 1 - sin x

cos² x = 1 - 2 sin x + sin² x

1 - sin² x = 1 - 2 sin x + sin² x

0 = - 2 sin x + 2 sin² x

0 = -2 sin x (1 - sin x)

sin x = 0 → x = 0

or

sin x = 1 → π / 2