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

sinx + cosx = 2

sinx -2 = -cosx = -(sqr(1-sin^2)

square both sides

sin^2x -4sinx + 4 = 1-sin^2

2sin^2x -4sinx + 3= 0

sin^2x -2xinx +3/2 = 0

look at the discriminant b^2 - 4ac = 4 -4(3/2) = 4-6 = -2

there is no real solution. There are imaginary solutions

The problem has sinx + cosx = 2. Maximum values for sinx and cosx are 1 each. At most they could sum to 2, but when sinx = 1, cosx doesn't = 1, so their sum could never be 2.

Answer 2

Since both sine and cosine are never greater than 1 the equation does not have a solution.