Find an equation of the tangent line to the graph of y =sin^2x at the point where x = pi/3

i understand the derivative is 2sinxcosx but after that i m stuffed please help

dy/dx = 2sin(x)·cos(x)
slope at (π/3, 3/4) = 2sin(π/3)cos(π/3) = 2((√3)/2)(1/2) = (√3)/2

Equation of line of slope (√3)/2 that passes through (π/3, 3/4):
y - 3/4 = ((√3)/2)(x - π/3)

dy/dx = 2sin(x)cos(x) = sin(2x)

Slope of tangent line at x = π/3 => sin(2π/3) = √3/2

Eqn of tangent line:

y - 3/4 = √3/2 * (x - π/3)

y = sin^2 x

dy/dx = 2 sin x cos x = sin 2x

At x = π/3, dy/dx = sin (2π/3) = √3/2

(π/3, 3/4)

y - 3/4 = (√3/2) (x - π/3)