The polynomial \(p(x)=ax^3+17x^2+bx-8\) is divisible by \(2x-1\) and has a remainder of \(-35\) when divided by \(x+3\).

(i) By finding the value of each of the constants \(a\) and \(b\), verify that \(a=b\).

Using your values of \(a\) and \(b\),

(ii) find \(p(x)\) in the form \((2x-1)q(x)\), where \(q(x)\) is a quadratic expression.

(iii) factorise \(p(x)\) completely.

(iv) solve \(a\sin^3\theta+17\sin^2\theta+b\sin\theta-8=0\) for \(0^\circ\lt \theta\lt 180^\circ\).