4cos^2 theta = 3+3 sin theta

S
=
{
209.551
...
+
360
∘
n
,
76.708
...
+
360
∘
n
}

Explanation:
Use the following formulas to transform the equation then solve
A
cos
x
+
B
sin
x
=
C
cos
(
x
–
D
)

C
=
√
A
2
+
B
2
,
cos
D
=
A
C
and
sin
D
=
B
C

A
=
−
4
,
B
=
3
,
C
=
√
16
+
9
=
√
25
=
5

cos
D
=
−
4
5
→
D
=
cos
−
1
(
−
4
5
)
=
143.13
...
∘
-> Quadrant II
5
cos
(
θ
−
143.13
...
∘
)
=
2

cos
(
θ
−
143.13
...
∘
)
=
2
5

θ
−
143.13
...
∘
=
cos
−
1
(
2
5
)

θ
−
143.13
...
∘
=
±
66.4218
...
+
360
∘
n

θ
=
143.13
...
±
66.4218
...
+
360
∘
n

θ
=
209.551
...
+
360
∘
n
,
76.708
...
+
360
∘
n

S
=
{
209.551
...
+
360
∘
n
,
76.708
...
+
360
∘
n
}