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NIMCET 2011 Question Paper with solutions
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© examsnet.com
Question : 15
Total: 120
The general value of
θ
satisfying the equation
2
sin
2
θ
−
3
sin
θ
−
2
=
0
, is
m
π
+
(
−
1
)
n
π
6
n
π
+
(
−
1
)
n
π
2
n
π
+
(
−
1
)
n
5
π
6
n
π
+
(
−
1
)
n
7
π
6
Validate
Solution:
Given,
2
sin
2
θ
−
3
sin
θ
−
2
=
0
⇒
2
sin
2
θ
−
4
sin
θ
+
sin
θ
−
2
=
0
⇒
2
sin
θ
⟨
sin
θ
−
2
+
1
(
sin
θ
−
2
)
=
0
⇒
(
sin
θ
−
2
)
(
2
sin
θ
+
1
)
=
0
⇒
sin
θ
=
−
1
2
,
sin
θ
≠
2
(
∵
−
1
≤
sin
θ
≤
1
)
⇒
θ
=
n
π
+
(
−
1
)
n
(
7
π
6
)
© examsnet.com
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