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Problems
Let
A
=
{
θ
∈
[
0
,
2
π
]
:
1
+
10
Re
(
2
cos
θ
+
i
sin
θ
cos
θ
−
3
i
sin
θ
)
=
0
}
A = \{\theta \in [0, 2\pi] : 1 + 10 \operatorname{Re}\left(\frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta}\right) = 0\}
A
=
{
θ
∈
[
0
,
2
π
]
:
1
+
10
Re
(
c
o
s
θ
−
3
i
s
i
n
θ
2
c
o
s
θ
+
i
s
i
n
θ
)
=
0
}
…
Mathematics
Problem
Let
A
=
{
θ
∈
[
0
,
2
π
]
:
1
+
10
Re
(
2
cos
θ
+
i
sin
θ
cos
θ
−
3
i
sin
θ
)
=
0
}
A = \{\theta \in [0, 2\pi] : 1 + 10 \operatorname{Re}\left(\frac{2\cos\theta + i\sin\theta}{\cos\theta - 3i\sin\theta}\right) = 0\}
A
=
{
θ
∈
[
0
,
2
π
]
:
1
+
10
Re
(
c
o
s
θ
−
3
i
s
i
n
θ
2
c
o
s
θ
+
i
s
i
n
θ
)
=
0
}
. Then
∑
θ
∈
A
θ
2
\sum_{\theta \in A} \theta^2
∑
θ
∈
A
θ
2
is equal to
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