JEE Main 2020Trigonometric EquationsTrigonometric EquationsHardQuestionLet S={θ∈[−π,π]−{± π2}:sinθtanθ+tanθ=sin2θ}S = \left\{ {\theta \in [ - \pi ,\pi ] - \left\{ { \pm \,\,{\pi \over 2}} \right\}:\sin \theta \tan \theta + \tan \theta = \sin 2\theta } \right\}S={θ∈[−π,π]−{±2π}:sinθtanθ+tanθ=sin2θ}. If T=∑θ ∈ Scos2θT = \sum\limits_{\theta \, \in \,S}^{} {\cos 2\theta } T=θ∈S∑cos2θ, then T + n(S) is equal to :OptionsA7 + 3\sqrt 3 3B9C8 + 3\sqrt 3 3D10Check AnswerHide SolutionSolutiontanθ(sinθ+1)−sin2θ=0\tan \theta(\sin \theta+1)-\sin 2 \theta=0tanθ(sinθ+1)−sin2θ=0 tanθ(sinθ+1−2cos2θ)=0⇒tanθ=0 or 2sin2θ+sinθ−1=0⇒(2sinθ+1)(sinθ−1)=0⇒sinθ=−12 or 1\begin{aligned} &\tan \theta\left(\sin \theta+1-2 \cos ^{2} \theta\right)=0 \\\\ &\Rightarrow \tan \theta=0 \text { or } 2 \sin ^{2} \theta+\sin \theta-1=0 \\\\ &\Rightarrow(2 \sin \theta+1)(\sin \theta-1)=0 \\\\ &\Rightarrow \sin \theta=\frac{-1}{2} \text { or } 1 \end{aligned}tanθ(sinθ+1−2cos2θ)=0⇒tanθ=0 or 2sin2θ+sinθ−1=0⇒(2sinθ+1)(sinθ−1)=0⇒sinθ=2−1 or 1 But, sinθ=1\sin \theta=1sinθ=1 not possible θ=0,π,−π,−π6,−5π6\theta=0, \pi,-\pi,-\frac{\pi}{6}, \frac{-5 \pi}{6}θ=0,π,−π,−6π,6−5π n(S)=5\mathrm{n}(\mathrm{S})=5n(S)=5 T=∑cos2θ=cos0∘+cos2π+cos(−2π)+cos(−5π3)+cos(−π3)T=\sum \cos 2 \theta=\cos 0^{\circ}+\cos 2 \pi+\cos (-2 \pi) +\cos \left(-\frac{5 \pi}{3}\right)+\cos \left(-\frac{\pi}{3}\right)T=∑cos2θ=cos0∘+cos2π+cos(−2π)+cos(−35π)+cos(−3π) = 4