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JEE Main 4 Apr 2024 Shift 1 Solved Paper
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© examsnet.com
Question : 7
Total: 90
If the system of equations
x
+
(
√
2
s
i
n
α
)
y
+
(
√
2
cos
α
)
z
=
0
x
+
(
cos
α
)
y
+
(
s
i
n
α
)
z
=
0
x
+
(
s
i
n
α
)
y
−
(
cos
α
)
z
=
0
has a non-trivial solution, then
α
∈
(
0
,
π
2
)
is equal to :
[4 Apr 2024 Shift 1]
3
π
4
7
π
24
5
π
24
11
π
24
Validate
Solution:
👈: Video Solution
|
1
√
2
s
i
n
α
√
2
cos
α
1
s
i
n
α
−
cos
α
1
cos
α
s
i
n
α
|
=
0
⇒
1
−
√
2
s
i
n
α
(
s
i
n
α
+
cos
α
)
+
√
2
cos
α
(
cos
α
−
s
i
n
α
)
=
0
⇒
1
+
√
2
cos
2
α
−
√
2
s
i
n
2
α
=
0
cos
2
α
−
s
i
n
2
α
=
−
1
√
2
cos
(
2
α
+
π
4
)
=
−
1
2
2
α
+
π
4
=
2
n
π
±
2
π
3
α
+
π
8
=
n
π
±
π
3
n
=
0
x
=
π
3
−
π
8
=
5
π
24
© examsnet.com
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