Find the smallest and largest value of the function f (x) = 2sinx + sin2x on the interval [pi / 2; pi]

Find the derivative of the function:

y ‘= (2 * sin (x) + sin (2x))’ = -2 * cos (x) – 2 * cos (2x) = -2 * cos (x) – 2 * (cos ^ 2 (x) – sin ^ 2 (x)) = -2 * cos (x) -2 * (2 * cos ^ 2 (x) – 1).

Let’s equate it to zero:

2 * cos ^ 2 (x) – cos (x) + 1 = 0

cos (x) = (1 + -√1 + 4 * 2 * 1) / 4 = (1 + -3) / 4

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

x1 = π / 2 x2 = 5π / 6

The root x2 does not belong to the given interval.

y (π / 2) = 2 * sin (π / 2) + sin (π) = 2

y (π) = 2 * sin (π) + sin (2π) = 0

Answer: 2 is the maximum value, 0 is the minimum value of the function.