xsinx-cosx=0 findx?
Answers
Answered by
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Step-by-step explanation:
Answer:
The answer is
S
=
{
2
k
π
,
π
2
+
2
k
π
}
,
k
∈
Z
Explanation:
We need
sin
(
A
+
B
)
=
sin
A
cos
B
+
sin
B
cos
A
sin
2
A
+
cos
2
A
=
1
We compare this equation to
r
sin
(
x
+
a
)
=
1
r
sin
x
cos
a
+
r
cos
x
sin
a
=
1
sin
x
+
cos
x
=
1
Therefore,
r
cos
a
=
1
and
r
sin
a
=
1
So,
cos
2
a
+
sin
2
a
=
1
r
2
+
1
r
2
=
2
r
2
=
1
r
2
=
2
,
⇒
,
r=sqrt2#
and
tan
a
=
1
,
⇒
,
a
=
π
4
Therefore,
√
2
sin
(
x
+
π
4
)
=
1
sin
(
x
+
π
4
)
=
1
√
2
x
+
π
4
=
π
4
+
2
k
π
,
⇒
,
x
=
2
k
π
and
x
+
π
4
=
3
π
4
+
2
k
π
,
⇒
,
x
=
π
2
+
2
k
π
The solutions are
S
=
{
2
k
π
,
π
2
+
2
k
π
}
,
k
∈
Z
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