sinx+sin2x+sin3x+sin4x=0
Answers
Answered by
0
Explanation:
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Here's the answer dear
In this way, using the sum-to-product formula for sinus and cosine:
sin
α
+
sin
β
=
2
sin
(
α
+
β
2
)
cos
(
α
−
β
2
)
cos
α
+
cos
β
=
2
cos
(
α
+
β
2
)
cos
(
α
−
β
2
)
.
So:
(
sin
4
x
+
sin
x
)
+
(
sin
3
x
+
sin
2
x
)
=
0
⇒
2
sin
(
4
x
+
x
2
)
cos
(
4
x
−
x
2
)
+
+
2
sin
(
3
x
+
2
x
2
)
cos
(
3
x
−
2
x
2
)
=
0
2
sin
(
5
2
x
)
cos
(
3
2
x
)
+
2
sin
(
5
2
x
)
cos
(
1
2
x
)
=
0
2
sin
(
5
2
x
)
[
cos
(
3
2
x
)
+
cos
(
1
2
x
)
]
=
0
2
sin
(
5
2
x
)
2
cos
(
3
2
x
+
1
2
x
2
)
cos
(
3
2
x
−
1
2
x
2
)
=
0
4
sin
(
5
2
x
)
cos
x
cos
(
x
2
)
=
0
.
Then:
sin
(
5
2
x
)
=
0
⇒
5
2
x
=
k
π
⇒
x
=
2
5
k
π
,
cos
x
=
0
⇒
x
=
π
2
+
k
π
,
Answered by
1
Answer:
sinx + sin2x + sin3x +sin4x = 0
Taking sin common:-
sin( x + 2x + 3x + 4x ) = 0
sin 10x = 0
As sin 0° = 0
⇒ sin 10x = sin 0°
10x = 0 {As sin and sin get cancelled out}
x = 0/10
x = 0
Explanation:
Hope it helps you :)
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