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Problem
Problem
The equation
2
sin
3
θ
+
(
2
λ
−
3
)
sin
2
θ
−
(
3
λ
+
2
)
sin
θ
−
2
λ
=
0
2 \sin^3 \theta + (2\lambda - 3) \sin^2 \theta - (3\lambda + 2) \sin \theta - 2\lambda = 0
2
sin
3
θ
+
(
2
λ
−
3
)
sin
2
θ
−
(
3
λ
+
2
)
sin
θ
−
2
λ
=
0
has exactly three roots in
(
0
,
2
π
)
(0, 2\pi)
(
0
,
2
π
)
, then
λ
\lambda
λ
can be equal to a. 0 b.
1
2
\frac{1}{2}
2
1
c. 1 d.
−
1
-1
−
1
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