(a) Solve \(6\sin^2x-13\cos x=1\) for \(0^\circ\leq x\leq360^\circ\).

(b) (i) Show that, for \(-\dfrac{\pi}{2}\lt y\lt \dfrac{\pi}{2}\), \(\dfrac{4\tan y}{\sqrt{1+\tan^2y}}\) can be written in the form \(a\sin y\), where \(a\) is an integer.

(ii) Hence solve \(\dfrac{4\tan y}{\sqrt{1+\tan^2y}}+3=0\) for \(-\dfrac{\pi}{2}\lt y\lt \dfrac{\pi}{2}\) radians.