find an equation of the tangent line to the curve y=3sinx-3sin^2x at the point (0,0).?

Answer 1

Find the derivative and solve for x=0.

dy/dx = 3cos(x)-18cos(x)sin(x)

dy/dx at x=0 is 3

Answer 2

Find the derivative, plug 0 for x. then plug your slope along with (0,0) in

y=mx+b

then you have b, its just a matter of rewriting it with the slope and the b you find

Answer 3

f `(x) = 3.cosx - 6sinx.cosx
f `(0) = 3 = m
Equation of tangent at (0,0) is given by:-
y = 3x