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Let’s check out your problem:
solve for
x
x
x
cos
x
+
sin
x
=
2
\cos x +\sin x = \sqrt{2}
cos
x
+
sin
x
=
2
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Math Problems
Algebra 1
Domain and range of square root functions: equations
Full solution
Q.
solve for
x
x
x
cos
x
+
sin
x
=
2
\cos x +\sin x = \sqrt{2}
cos
x
+
sin
x
=
2
Recognize Equality Condition:
Recognize that
cos
(
x
)
\cos(x)
cos
(
x
)
and
sin
(
x
)
\sin(x)
sin
(
x
)
are equal when
x
x
x
is
45
45
45
degrees or
π
4
\frac{\pi}{4}
4
π
radians.
Use Trigonometric Identity:
Use the identity
sin
(
x
)
=
cos
(
π
2
−
x
)
\sin(x) = \cos(\frac{\pi}{2} - x)
sin
(
x
)
=
cos
(
2
π
−
x
)
to rewrite the equation as
cos
(
x
)
+
cos
(
π
2
−
x
)
=
2
\cos(x) + \cos(\frac{\pi}{2} - x) = \sqrt{2}
cos
(
x
)
+
cos
(
2
π
−
x
)
=
2
.
Substitute
x
x
x
Value:
Since
cos
(
x
)
=
cos
(
π
2
−
x
)
\cos(x) = \cos(\frac{\pi}{2} - x)
cos
(
x
)
=
cos
(
2
π
−
x
)
when
x
=
π
4
x = \frac{\pi}{4}
x
=
4
π
, substitute
π
4
\frac{\pi}{4}
4
π
into the equation to check if it satisfies the equation.
Calculate Cosine and Sine:
Calculate
cos
(
π
4
)
+
sin
(
π
4
)
\cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4})
cos
(
4
π
)
+
sin
(
4
π
)
and see if it equals
2
\sqrt{2}
2
.
\newline
cos
(
π
4
)
=
2
2
\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}
cos
(
4
π
)
=
2
2
and
sin
(
π
4
)
=
2
2
.
\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}.
sin
(
4
π
)
=
2
2
.
Add Values:
Add the values:
2
/
2
+
2
/
2
=
2
\sqrt{2}/2 + \sqrt{2}/2 = \sqrt{2}
2
/2
+
2
/2
=
2
.
Verify Solution:
Since the sum equals
2
\sqrt{2}
2
,
x
=
π
4
x = \frac{\pi}{4}
x
=
4
π
is a solution.
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\newline
2
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Question
lim
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2
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x
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3
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lim
x
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3
x
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2
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Question
lim
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1
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lim
x
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Question
lim
x
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1
x
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1
x
+
5
−
2
=
\lim _{x \rightarrow-1} \frac{x+1}{\sqrt{x+5}-2}=
lim
x
→
−
1
x
+
5
−
2
x
+
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=
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Question
Let
f
(
x
)
=
x
+
9
f(x)=\sqrt{x+9}
f
(
x
)
=
x
+
9
and let
c
c
c
be the number that satisfies the Mean Value Theorem for
f
f
f
on the interval
[
0
,
16
]
[0,16]
[
0
,
16
]
.
\newline
What is
c
c
c
?
\newline
Choose
1
1
1
answer:
\newline
(A)
16
16
16
\newline
(B)
43
43
43
\newline
(C)
55
55
55
\newline
(D)
7
7
7
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Question
Let
g
(
x
)
=
5
x
−
1
g(x)=\sqrt{5 x-1}
g
(
x
)
=
5
x
−
1
and let
c
c
c
be the number that satisfies the Mean Value Theorem for
g
g
g
on the interval
[
1
,
10
]
[1,10]
[
1
,
10
]
.
\newline
What is
c
c
c
?
\newline
Choose
1
1
1
answer:
\newline
(A)
2
2
2
.
25
25
25
\newline
(B)
4
4
4
.
25
25
25
\newline
(C)
6
6
6
.
5
5
5
\newline
(D)
8
8
8
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