Solve sin x − cos x = 1 if 0 ≤ x ≤ 2π.

Answer 1

A fairly straight forward method is the square both sides of the equation. However, in doing so, we will be introducing a potential "non solution" condition so we will need to review the answers we get to ensure they are all valid.

sin(x) − cos(x) = 1

(sin(x) − cos(x))² = 1²

sin²(x) - 2sin(x)cos(x) + cos²(x) = 1

sin²(x) + cos²(x) - 2sin(x)cos(x) = 1

Using the Pythagorean Identity sin²(x) + cos²(x) = 1:

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

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

sin(x)cos(x) = 0

So either sin(x) = 0 (meaning x = 0, π, and 2π) or cos(x) = 0 (meaning x = π/2 and 3π/2).

Checking our answers:

For x = 0: sin(0) - cos(0) = 1 is NOT true

For x = π: sin(π) - cos(π) = 1 is TRUE

For x = 2π: sin(2π) - cos(2π) = 1 is NOT true

For x = π/2: sin(π/2) - cos(π/2) = 1 is TRUE

For x = 3π/2: sin(3π/2) - cos(3π/2) = 1 is NOT true

So the answer turns out to be x = π/2 and x = π