Let A= θ ∈[0,2 π]: 1+10 Re ( 2 cos θ+i sin θ cos θ-3 i sin θ )=0 . Then Σ_θ ∈ A θ^2 is equal to

$$\begin{aligned} & 1+10 \operatorname{Re}\left(\frac{2 \cos \theta+i \sin \theta}{\cos \theta-3 i \sin \theta}\right)=0 \\ & \therefore \mathrm{z}+\overline{\mathrm{z}}=2 \operatorname{Re}(\mathrm{z}) \\ & \frac{2 \cos \theta+\mathrm{i} \sin \theta}{\cos \theta-3 \mathrm{i} \sin \theta}+\frac{2 \cos \theta-\mathrm{i} \sin \theta}{\cos \theta+3 \mathrm{i} \sin \theta}=2 \times\left(\frac{-1}{10}\right) \\ & \frac{\left(2 \cos ^2 \theta-3 \sin ^2 \theta\right)+\left(2 \cos ^2 \theta\right)-\left(3 \sin ^2 \theta\right)}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-2}{10} \\ & \Rightarrow \frac{2 \cos ^2 \theta-3 \sin ^2 \theta}{\cos ^2 \theta+9 \sin ^2 \theta}=\frac{-1}{10} \\ & \Rightarrow 20 \cos ^2 \theta-30 \sin ^2 \theta=-\cos ^2 \theta-9 \sin ^2 \theta \\ & 21 \cos ^2 \theta-21 \sin ^2 \theta=0 \\ & \Rightarrow \cos 2 \theta=0 \\ & 2 \theta=\frac{\pi}{2}, \frac{3 \pi}{2}, \frac{5 \pi}{2}, \frac{7 \pi}{2} \\ & \Rightarrow \sum \theta^2=\frac{\pi^2}{16}+\frac{9 \pi^2}{16}+\frac{25 \pi^2}{16}+\frac{49 \pi^2}{16}=\frac{84 \pi^2}{16}=\frac{21 \pi^2}{4}\end{aligned}$$​1+10Re(cosθ−3isinθ2cosθ+isinθ​)=0∴z+z=2Re(z)cosθ−3isinθ2cosθ+isinθ​+cosθ+3isinθ2cosθ−isinθ​=2×(10−1​)cos2θ+9sin2θ(2cos2θ−3sin2θ)+(2cos2θ)−(3sin2θ)​=10−2​⇒cos2θ+9sin2θ2cos2θ−3sin2θ​=10−1​⇒20cos2θ−30sin2θ=−cos2θ−9sin2θ21cos2θ−21sin2θ=0⇒cos2θ=02θ=2π​,23π​,25π​,27π​⇒∑θ2=16π2​+169π2​+1625π2​+1649π2​=1684π2​=421π2​​