(a) Clear the denominator, replace \(\tan x=\sin x/\cos x\), and multiply by \(\cos x\): \[\sin x+\cos^2x=k(\sin x-\cos^2x).\] Use \(\cos^2x=1-\sin^2x\) and collect terms: \[(k+1)\sin^2x+(k-1)\sin x-(k+1)=0,\] as required. These steps apply wherever the original expression is defined.

(b) With \(k=4\), \[5\sin^2x+3\sin x-5=0,\qquad \sin x=\frac{-3\pm\sqrt{109}}{10}.\] Reject the negative root, which is less than \(-1\). The positive root is \(0.7440306\ldots\), giving \[\boxed{x=48.1^\circ,\quad131.9^\circ}.\] Both lie in the interval and satisfy the original equation with nonzero denominator.