Find the x-coordinates of all points on the graph of: f(x)=6sinx-sin^2x
at which the tangent line is horizontal.

The points where the tangent line of f(x) are horizontal are those where f '(x) = 0.

Then find f '(x)

f '(x) = 6cosx - 2sinxcosx = 2cosx (3 - sinx) =0

=> 3 - sinx = 0, which is never

or cosx = 0, which is for x = π/2, 3π/2, 5π2, , -π/2 , -3π/2, -5π/2 , ...

You can represent that by x = π/2 + kπ, for k ∈ Z.
 

Cis your correct answer