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JEE Main Math Class 11 Trigonometric Functions Questions
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© examsnet.com
Question : 2
Total: 115
Suppose
θ
∈
[
0
,
π
4
]
is a solution of
4
cos
θ
−
3
s
i
n
θ
=
1
.
Then
cos
θ
is equal to :
[5 Apr 2024 Shift 1]
4
(
3
√
6
−
2
)
6
−
√
6
(
3
√
6
−
2
)
6
+
√
6
(
3
√
6
+
2
)
4
(
3
√
6
+
2
)
Validate
Solution:
👈: Video Solution
4
(
1
−
tan
2
θ
∕
2
1
+
tan
2
θ
∕
2
)
−
3
(
2
tan
θ
2
1
+
tan
2
θ
2
)
=
1
let
tan
θ
2
=
t
4
−
4
t
2
−
6
t
1
+
t
2
=
1
4
−
4
t
2
−
6
t
=
1
+
t
2
⇒
5
t
2
+
6
t
−
3
=
0
⇒
t
=
−
6
±
√
36
−
4
(
5
)
(
−
3
)
2
(
5
)
=
−
6
±
√
96
10
=
−
6
±
4
√
6
10
t
=
−
3
+
2
√
6
5
cos
θ
=
1
−
t
2
1
+
t
2
=
1
−
(
2
√
6
−
3
5
)
2
1
+
(
2
√
6
−
3
5
)
2
=
1
−
(
24
+
9
−
12
√
6
25
)
1
+
(
24
+
9
−
12
√
6
25
)
=
25
−
33
+
12
√
6
25
+
33
−
12
√
6
=
12
√
6
−
8
58
−
12
√
6
=
6
√
6
−
4
29
−
6
√
6
×
29
+
6
√
6
29
+
6
√
6
=
100
+
150
√
6
625
=
4
+
6
√
6
25
×
4
−
6
√
6
4
−
6
√
6
=
−
200
25
(
4
−
6
√
6
)
=
−
8
4
−
6
√
6
=
4
3
√
6
−
2
© examsnet.com
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