F (x) = 9 sin (x) + cot (x), - π ≤ x ≤ π find the interval of increase.

We are given the function

F (x) = 9 sinx + cot x

We need to take the first derivative of the given function so,

F' (x) = 9 cos x - csc² x

Next, we equate the first derivative of the function to 0 and solve for the values of x

0 = 9 cos x - csc² x

Solving for x

x = 2.04

Picking out an arbitrary value between 2.04 and π, say 3 and substituting in F (x)

F (3) = 9 sin 3 + cot 3 = 19.55

Therefore, the interval where the function is increasing is from 2.04 to π

Consequently, the interval where the function is decreasing is from - π to 2.04