Let S = \ θ ∈ [ - π, π ] - \ ± (π)/(2)\ : sinθ tanθ + tanθ = 2θ\ ....

Given $sin\theta tan\theta + tan\theta = \sin 2\theta; \theta \ne \pm \dfrac{\pi}{2}$

$⇒ \dfrac{\sin^{2}\theta}{cos\theta} + \dfrac{sin\theta}{cos\theta} = 2sin\theta cos\theta$

$⇒ sin\theta = 0$ or $\dfrac{sin\theta}{cos\theta} + \dfrac{1}{cos\theta} = 2cos\theta$

$⇒ \theta = 0, \pi, - \pi$ or $sin\theta + 1 = 2\left(1 - \sin^{2}\theta\right)$

$⇒ \theta = 0, \pi, - \pi$ or $2\sin^{2}\theta + sin\theta - 1 = 0$

$⇒ \theta = 0, \pi, - \pi$ or $sin\theta = \dfrac{1}{2}, sin\theta = - 1$

$⇒ \theta = 0, \pi, - \pi$ or $\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$ or $\theta = \dfrac{- \pi}{2}$ (rejected)

Hence, $\theta = \left\{0, \pi, - \pi, \dfrac{\pi}{6}, \dfrac{5\pi}{6}\right\}$

$⇒ n\left(S\right) = 5$

Now, $T = Σ\cos 2\theta$

$= \cos 0 + \cos 2\pi + \cos \left(- 2\pi\right) + \cos \left(\dfrac{\pi}{3}\right) + \cos \left(\dfrac{5\pi}{3}\right)$

$= 1 + 1 + 1 + \dfrac{1}{2} + \dfrac{1}{2} = 4$

So, $T + n\left(S\right) = 9$