Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.) sin x + cos x = 2

Question

Solve the equation for x if 0 ≤ x < 2π. Give your answer in radians using exact values only. (Enter your answers as a comma-separated list.)

sin x + cos x = 2

Answer 1

sin(x) + cos(x) = √2

(sin(x) + cos(x))² = (√2)²

sin²(x) + cos²(x) + 2sin(x)cos(x) = 2

1 + sin(2x) = 2

sin(2x) = 1

x = (1/2) sin^-1(1)

x = (1/2) (π/2, 3π/2, 5π/2, 7π/2, 9π/2...)

x = π/4, 3π/4, 5π/4, 7π/4, 9π/4 ...

For 0 ≤ x < 2π, we have x = π/4, 3π/4, 5π/4, 7π/4. However, plugging in these values, we see that only x = π/4 is valid. Therefore, the solution is

x = π/4