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Byju's Answer
Standard XII
Mathematics
Properties Derived from Trigonometric Identities
If the equati...
Question
If the equation
sin
4
x
−
(
k
+
2
)
sin
2
x
−
(
k
+
3
)
=
0
has a solution then
k
must lie in the interval:
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Solution
s
i
n
4
x
−
(
k
+
2
)
s
i
n
2
x
−
(
k
+
3
)
=
0
s
i
n
2
x
∈
[
0
,
1
]
put
s
i
n
2
x
=
t
,
t
∈
[
0
,
1
]
t
2
−
(
k
+
2
)
t
−
(
k
+
3
)
=
0
put
t
=
0
0
−
(
k
+
2
)
0
−
k
−
3
=
0
k
=
−
3
put
t
=
1
1
−
k
−
2
−
k
−
3
=
0
⇒
−
2
k
=
4
k
=
−
2
k
∈
[
−
3
,
−
2
]
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0
Similar questions
Q.
The equation
sin
4
x
−
(
k
+
2
)
sin
2
x
−
(
k
+
3
)
=
0
possesses a solution if
Q.
If the equation
sin
4
x
−
(
k
+
2
)
sin
2
x
−
(
k
+
3
)
=
0
has a solution then k must lie in the interval
Q.
If the equation
sin
4
x
−
(
k
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2
)
sin
2
x
−
(
k
+
3
)
=
0
has a solution, then
k
must lie in the interval
Q.
If the equation
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+
24
x
+
k
=
0
has exactly one root in
(
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