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MHT CET 2019 Math Solved Paper
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© examsnet.com
Question : 40
Total: 50
The number of solutions of
sin
2
θ
=
1
2
i
n
[
0
,
π
]
is ..........
three
four
two
one
Validate
Solution:
∵
sin
2
q
=
1
2
⇒
sin
2
θ
=
(
1
√
2
)
2
⇒
sin
2
θ
=
sin
2
(
π
4
)
⇒
θ
=
n
π
±
π
4
,
n
∊
I
∴ in [0,π] , there are only two solutions i.e ;
π
4
and
3
π
4
© examsnet.com
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