Let f be an odd function defined on the set of real numbers such that for x≥0, f(x) = 3sin x + 4cos x. Then f(x) at x = 11π/6 is equal to:

f(x) = 3sin x + 4cos x x≥0
f(x) = 5sin(x+α) α = sin⁻¹(4/5)
f(11π/6) = -f(11π/6) = -[3sin(11π/6) + 4cos(11π/6)] = -[3sin(2π - π/6) + 4cos(2π - π/6)] = -[-3/2 + 4√3/2] = 3/2 - 2√3 = 32 - √3