Here, \(\begin{array}{l} tan~\alpha =\sqrt{5}\end{array}\)
\(\begin{array}{l} \therefore\ \tan\theta=\frac{\tan\alpha-\tan2\theta}{1+\tan\alpha\tan2\theta} \end{array}\)
∴ tan θ = tan (α – 2θ)
\(α – 2θ = nπ + θ\)
⇒ \(3θ = α – nπ\)
\(\begin{array}{l} \Rightarrow\ \theta = \frac{\alpha}{3}-\frac{n\pi}{3}~;~n\in Z\end{array}\)
If θ [–π, π/2) then
n = 0, 1, 2, 3, 4 are acceptable
∴ 5 solutions.