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Byju's Answer
Standard XII
Mathematics
General Solution of sin theta = sin alpha
The general s...
Question
The general solution of the equation
7
cos
2
θ
+
3
sin
2
θ
=
4
is
(a)
θ
=
2
n
π
±
π
6
,
n
∈
Z
(b)
θ
=
2
n
π
±
2
π
3
,
n
∈
Z
(c)
θ
=
n
π
±
π
3
,
n
∈
Z
(d) none of these
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Solution
(c)
θ
=
n
π
±
π
3
,
n
∈
Z
Given:
7
cos
2
θ
+
3
sin
2
θ
=
4
⇒
7
cos
2
θ
+
3
(
1
-
cos
2
θ
)
=
4
⇒
7
cos
2
θ
+
3
-
3
cos
2
θ
=
4
⇒
4
cos
2
θ
+
3
=
4
⇒
4
(
1
-
cos
2
θ
)
=
3
⇒
4
sin
2
θ
=
3
⇒
sin
2
θ
=
3
4
⇒
sin
θ
=
3
2
⇒
sin
θ
=
sin
π
3
⇒
θ
=
n
π
±
π
3
,
n
∈
Z
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0
Similar questions
Q.
The general solution of the equation
7
cos
2
x
+
3
sin
2
x
=
4
is
(a)
x
=
2
n
π
±
π
6
,
n
∈
Z
(b)
x
=
2
n
π
±
2
π
3
,
n
∈
Z
(c)
x
=
n
π
±
π
3
,
n
∈
Z
(d) none of these
Q.
If
A
=
cos
θ
-
sin
θ
sin
θ
cos
θ
, then A
T
+ A = I
2
, if
(a) θ = n π, n ∈ Z
(b) θ = (2n + 1)
π
2
, n ∈ Z
(c) θ = 2n π +
π
3
, n ∈ Z
(d) none of these
Q.
If
4
sin
2
θ
=
1
, then the values of θ are
(a)
2
n
π
±
π
3
,
n
∈
Z
(b)
n
π
±
π
3
,
n
∈
Z
(c)
n
π
±
π
6
,
n
∈
Z
(d)
2
n
π
±
π
6
,
n
∈
Z
Q.
If
cot
θ
-
tan
θ
=
sec
θ
, then, θ is equal to
(a)
2
n
π
+
3
π
2
,
n
∈
Z
(b)
n
π
+
-
1
n
π
6
,
n
π
Z
(c)
n
π
+
π
2
,
n
∈
Z
(d) none of these.
Q.
If
4
sin
2
x
=
1
, then the values of
x
are
(a)
2
n
π
±
π
3
,
n
∈
Z
(b)
n
π
±
π
3
,
n
∈
Z
(c)
n
π
±
π
6
,
n
∈
Z
(d)
2
n
π
±
π
6
,
n
∈
Z
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