Objective:
To find the maximum and minimum value of the function f(x) = sin(x) + 1/2 * cos(2x) in the interval [0, π/2].

Solution:

Step 1: Find the critical points of the function by taking the derivative.

The derivative of f(x) is given by:
f'(x) = cos(x) - sin(2x)

To find the critical points, we need to solve the equation f'(x) = 0.
cos(x) - sin(2x) = 0

Step 2: Solve the equation to find the critical points.

cos(x) - sin(2x) = 0
cos(x) - (2sin(x)cos(x)) = 0 (using the identity sin(2x) = 2sin(x)cos(x))
cos(x)(1 - 2sin(x)) = 0

This equation will be satisfied if either cos(x) = 0 or (1 - 2sin(x)) = 0.

Case 1: cos(x) = 0

In the interval [0, π/2], cos(x) is equal to zero at x = π/2.

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

Solving for sin(x):
1 - 2sin(x) = 0
2sin(x) = 1
sin(x) = 1/2
x = π/6 or x = 5π/6

Therefore, the critical points are x = π/2, π/6, and 5π/6.

Step 3: Evaluate the function at the critical points and the endpoints of the interval [0, π/2].

f(0) = sin(0) + 1/2 * cos(0) = 0 + 1/2 * 1 = 1/2
f(π/2) = sin(π/2) + 1/2 * cos(2(π/2)) = 1 + 1/2 * 0 = 1

f(π/6) = sin(π/6) + 1/2 * cos(2(π/6)) = 1/2 + 1/2 * (√3/2) = 1/2 + √3/4
f(5π/6) = sin(5π/6) + 1/2 * cos(2(5π/6)) = 1/2 + 1/2 * (-√3/2) = 1/2 - √3/4

Step 4: Compare the values obtained in Step 3 to find the maximum and minimum values.

The maximum value of f(x) in the interval [0, π/2] is 1, which occurs at x = π/2.
The minimum value of f(x) in the interval [0, π/2] is 1/2 - √3/4, which occurs at x = 5π/6.

Therefore, the maximum value of f(x) is 1 and the minimum value is 1/2 -