(a) Let \(q=\tan x\). Using \(\tan2x=2q/(1-q^2)\), \[q^3+\frac{4q}{1-q^2}-q=0.\] Multiply by \(1-q^2\), which is nonzero wherever the original equation is defined: \[-q^5+2q^3+3q=0.\] As q≠0, divide by −q to obtain \(q^4-2q^2-3=0\).

(b) First retain the case excluded by division: \(\tan(2\theta)=0\). In (0,π), this gives \(\theta=\pi/2\), and it satisfies the original equation. For nonzero \(\tan(2\theta)\), \[(q^2-3)(q^2+1)=0,\quad q=\tan(2\theta)=\pm\sqrt3.\] Hence all solutions are \[\boxed{\theta=\frac\pi6,\ \frac\pi3,\ \frac\pi2,\ \frac{2\pi}3,\ \frac{5\pi}6}.\] Each makes both tangents defined. The published MS lists the four nonzero-tangent solutions but omits the valid solution π/2; it is retained here.