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Byju's Answer
Standard XII
Mathematics
Equations with Solutions at Boundary Values
The number of...
Question
The number of solution of
sin
4
θ
−
2
sin
2
θ
−
1
=
0
for
θ
∈
[
0
,
2
π
]
is
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Solution
Given:
sin
4
θ
−
2
sin
2
θ
−
1
=
0
⇒
(
sin
2
θ
−
1
)
2
−
2
=
0
⇒
sin
2
θ
=
1
±
√
2
As
sin
2
θ
∈
[
0
,
1
]
, so no solution of the given equation.
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Equations with Solutions at Boundary Values
Standard XII Mathematics
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