a.maths trigo

f(x) = 2 sin x - x
f '(x) = 2 cos x - 1
when f '(x) = 0, x = π/3

f(0) = 2 sin 0 - 0 = 0
f(π) = 2 sin π - π = -π
f(π/3) = 2 sin (π/3) - (π/3) = (3√3 - π)/3

thus the greatest value is (3√3 - π)/3,
the least value is -π

why don't use f '(x)? it is the simpliest way to do!

The greatest value of f(x) = √3 - π/3 when x = π/3

The least value of f(x) = - π when x = π over the given 0<= x <= π