RANGE

4x² - 8x + 3
=> 4x² - 2x - 6x + 3
=> 2x(2x - 1) - 3(2x - 1)
=> (2x - 3)(2x - 1) ≤ 0

Our checkpoints are 3/2 and 1/2. Using wavy curve method, we get,
f(x) ≤ 0 for all x → [1/2, 3/2]

This implies θ → [sin^-1(1/2), sin^-1(3/2)]
θ → [pi/6, sin^-1(3/2)]
To find sin^-1(3/2), we see that 3/2 < pi

So sin^-1(3/2) = sin^-1(pi - 3/2)
= sin^-1((2pi - 3)/2)

Nothing else I could think of..

Edit :

(2sinθ - 3)(2sinθ - 1) ≤ 0
Now 2sinθ - 3 will always be negative as -1 ≤ sinθ≤ 1
So to get an overall negative result, we require that :
2sinθ - 1 ≥ 0
=> sinθ ≥ 1/2
=> θ ≥ π6

How it ends in π/3 I don't know..lol.