To solve the inequality √2sin(pi/2-x) < 1, we need to first simplify the expression:

√2sin(pi/2-x) = √2cos(x)

Now, our inequality becomes:

√2cos(x) < 1

Squaring both sides to eliminate the square root:

2cos(x) < 1

Dividing by 2:

cos(x) < 1/2

To solve for x, we need to find the angles in the interval [0, 2π] where cos(x) < 1/2.

The solutions to this inequality are in the first and fourth quadrants where the cosine function is positive. You can use the unit circle or a graphing calculator to find the angles where cos(x) < 1/2.

The solutions are:

x ∈ (π/3, 5π/3)

Therefore, the inequality √2sin(pi/2-x) < 1 is satisfied when x ∈ (π/3, 5π/3).