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Trigonometry |
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Trigonometric
Equations |
The
Equations,
sin
(bx + c) = m, -1
<
m <
1,
cos
(bx + c) = m,
-1
<
m <
1,
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tan
(bx
+ c) = m and cot
(bx
+ c) = m,
where
b,
c and
m are real
numbers. |
The
Equation
sin
(bx + c) = m, -1
<
m <
1, example |
The
Equation
cos
(bx + c) = m,
-1
<
m <
1, example |
The
Equation
tan
(bx
+ c) = m, example |
The
Equation
cot
(bx
+ c) = m, example |
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The Equations,
sin
(bx + c) = m, -1
<
m <
1,
cos
(bx + c) = m, -1
<
m <
1, |
tan
(bx
+ c) = m and cot
(bx
+ c) = m,
where
b,
c
and m
are real numbers. |
The given equations can be written as F
(bx +
c) = m
where F
substitutes a trigonometric function, x
is an arc
to be calculated and m
is a value of a given trigonometric function. |
To every trigonometric function we can determine an arc,
a +
k · P
of which function value equals
m
that is F
(a +
k · P) = m,
where a
= x0
is the basic solution, and P
is the period, then |
F
(bx +
c) = F (a +
k · P)
or bx +
c = a +
k · P,
thus |
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Thus, from obtained general solution we can write a common solutions for every given
equation, |
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The Equation
sin
(bx + c) = m, -1
<
m <
1, example |
Example: Solve
the equation, |
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Solution: Rewrite
the equation to the form sin
(bx + c) = m, so
sin
(2x + p/6)
= -
1/2
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An alternative but similar solution can be obtained by substituting the values of,
b,
c
and m,
into |
x0
= a
and x′0
= p
-
a
and to the common solution
written above |
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The Equation
cos
(bx + c) = m,
-1
<
m <
1, example |
Example: Find
the solutions of the equation, 2cos
(4x -
30°) + Ö3
= 0.
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Solution: Rewrite
the equation to the form cos
(bx + c) = m, that
is cos
(4 x -
30°) = -
Ö3/2
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it
follows that cos
(4 x -
30°)
= cos (± 150° +
k · 360°) |
and
4 x -
30°
= ± 150° +
k · 360° |
so,
x = 45° +
k · 90°
and x′ = -
30° +
k · 90°, kÎ
Z. |
The same results we get by substituting the values, b
= 4,
c = -
30°
and m =
-
Ö3/2,
into |
x0
= a
= cos-1
m
= cos-1
(-
Ö3/2)
= 150° and
x′0
= -
a
= -150° |
then,
using the common solution formulas obtained are |
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The Equation
tan
(bx
+ c) = m, example |
Example: Solve
the equation, |
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Solution: Rewrite
the equation to the form tan
(bx + c) = m,
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We obtain the same result if we put given parameters,
b,
c, and corresponding basic solution |
x0
= a
= tan-1
m to the common
solution that is, b
= 1/3,
c = -
p/2 and
x0
= a
= tan-1(-Ö3/3) =
-
p/6 |
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The Equation
cot
(bx
+ c) = m, example |
Example: Find
the solutions of the equation, cot
(-
2x + 10°) -
1 = 0.
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Solution: Rearrange
the given equation to the form cot
(bx + c) = m, thus
cot (-
2x + 10°) = 1,
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or cot
[-
(2x -
10°)] = 1
and since cot
(-
a)
= -
cot a
then, cot
(2x -
10°) = -
1, |
and
cot (2x
-
10°) = cot (135° + k · 180°),
2x -
10° = 135° + k · 180° => x = 72°30′ +
k · 90°. |
The general solution of the equation we get direct substituting the basic solution
x0
= a
and the constant b |
and
c
to the common solution, b
= 2,
c = -
10° and
x0
= a
= cot-1(-1) =
135°
give |
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Functions
contents D |
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