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| Trigonometry |
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Trigonometric
Equations |
Homogeneous Equations
in sin
x and cos
x |
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Homogeneous equations
of first degree a×sin x
+ b×cos
x = 0 |
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Homogeneous equations
of second degree a
sin2
x
+ b sin x
· cos
x + c
cos2
x = 0
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The
Basic Strategy for Solving Trigonometric Equations |
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Trigonometric equations
examples |
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| Homogeneous
trigonometric equations
in sin
x and cos
x |
| An equation is said to be homogeneous if all its terms are of the same degree.
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| Homogeneous equations
of first degree a sin x
+ b
cos
x = 0 |
| Divide given equation by
cos x to obtain |
| a
tan x + b =
0
or tan
x = -
b/a
the basic equation
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| whose solution is
x =
tan-1
(-
b/a) + k · p
or x
= arctan
(-
b/a) + k · p,
k Î Z. |
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Example: Solve the equation,
-
sin x
+
Ö3
· cos x = 0.
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Solution: Dividing given equation by
-
cos x
obtained is
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| tan
x =
Ö3,
x =
tan-1Ö3
+ k · p =
p/3
+ k · p,
k Î Z. |
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| Homogeneous equations
of second degree a
sin2
x
+ b sin x
· cos
x + c
cos2
x = 0 |
| After division of the given equation by
cos2
x obtained is quadratic
equation |
| a
tan2
x
+ b tan x + c = 0 |
| which is
explained in the previous section. |
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| The
next example shows that the equation a
sin2
x
+ b sin x
· cos
x + c
cos2
x = d
is also homogeneous. |
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Example: Find the solution set for the equation,
5
sin2
x
+ sin x · cos
x + 2
cos2
x = 4.
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Solution: Given equation is equivalent to the equation
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| 5
sin2
x
+ sin x · cos
x + 2
cos2
x = 4
· (sin2
x
+ cos2
x),
since sin2
x
+ cos2
x = 1 |
| which,
when simplified becomes |
| sin2
x
+ sin x · cos
x - 2
cos2
x = 0
- the homogeneous equation of the second degree. |
| Division by cos2
x
gives, |
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thus, (tan x)1
= - 2,
x =
tan-1(- 2)
= -
63°26′05″ + k · 180°,
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(tan x)2
= 1,
x′ =
tan-1
1
= 45° + k · 180°,
k Î Z. |
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| The
basic strategy for solving trigonometric equations |
| When solving trigonometric equations we usually use some of the following
procedures, |
| - apply known
identities to modify given equation to an equivalent expressed in terms of one function, |
| - rearrange the given
equation using different trigonometric formulae to an equivalent, until you recognize one |
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of the known types. |
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| Trigonometric equations
examples |
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Example: Solve the equation,
3
sin
(x
+ 70°) + 5
sin
(x
+ 160°) = 0.
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Solution: Given equation
can be written as
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| 4
sin
(x
+ 70°) - sin
(x
+ 70°) + 4
sin
(x
+ 160°)
+ sin
(x
+ 160°) = 0 |
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or
4
[sin
(x
+ 70°)
+ sin
(x
+ 160°)] =
sin
(x
+ 70°) - sin
(x
+ 160°) |
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| then, by using sum to product formula
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cot
(- 45°) · tan
(x
+ 115°) = 1/4
or tan
(x
+ 115°) = - 1/4,
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| therefore,
the solution x
+ 115° = tan-1
, x =
-115°
+ tan-1
(-1/4) =
- 129°2′10″ + k · 180°. |
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Example: Solve the equation,
cos
2x
+ cos
6x - cos
8x -
1 = 0.
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Solution: To the first two terms apply the sum to product formula and remaining two terms transform using
known identity, thus
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 |
| and
since 1
+ cos
2x = 2cos
2 x
then, 1
+ cos
8x = 2cos
2 4x plugging into the
given equation |
| 2cos
4x · cos
2x -
2cos
2 4x
= 0
or 2cos
4x · (cos
2x -
cos
4x)
= 0 |
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| it
follows that 4cos
4x · sin
3x · sin
x
= 0
meaning,
cos
4x
= 0,
sin
3x
= 0
and/or sin
x
= 0.
Since |
| sin
3x
= sin (x
+ 2x)
= sin
x · cos
2x + cos
x · sin
2x
= sin
x · (cos
2 x -
sin
2 x) + cos
x · 2sin
xcos
x |
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=
sin
x · (1 -
2sin
2 x) + 2sin
x · (1 -
sin
2 x)
=
3sin
x -
4sin
3 x,
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solutions of the equation
sin
x
= 0
are included in the solutions of sin
3x
= 0.
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If sin
x
= 0
then,
sin
3x
= 3sin
x -
4sin
3 x
= sin
x · (3 -
4sin
2 x)
= 0.
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Therefore, the solution set of the given equation is the union of the solutions of each of the equations, that is
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cos
4x
= 0,
4x
= ± p/2
+ k · 2p,
x
=
± p/8
+ k · p/2
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and sin
3x
= 0,
3x
= k · p,
x
= k · p/3,
k Î Z.
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| Trigonometry
contents B |
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