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| Trigonometry |
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Trigonometric
Equations |
Trigonometric
Equations of Quadratic Form |
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Equations
of the Type
a
· cos
x +
b
· sin
x = c |
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Introducing new unknown
t
= tan x/2 |
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Introducing new unknown
t
= tan x/2
example |
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The
Basic Strategy for Solving Trigonometric Equations |
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Trigonometric
equations examples |
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| Trigonometric
equations of quadratic form |
| The trigonometric equation of the quadratic form
[F (x)]2
+ p · F (x) + q = 0, where
F (x)
denotes given trigonometric function, by substituting F
(x) = u becomes a quadratic equation |
 |
| returned into substitutions
F
(x) = u1
and F
(x) = u2
lead to the known basic trigonometric equation. |
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Example: Find the solution set for the equation,
3sin x = 2cos2
x.
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Solution: Using known identity
we write 3sin
x -
2 · (1 -
sin2 x) = 0,
and by plugging
sin x =
u
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| Therefore,
sin x =
1/2,
x = 30° + k · 360°, kÎ Z,
and x′ = 150° + k · 360°, kÎ Z,
while the equation sin x =
-
2 has no solutions since
-
2 is not in the range of the sine function. |
| Thus,
the solution set of the given equation is {30° + k · 360°, 150° + k · 360° , kÎ
Z}. |
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| Equations of the type
a
cos
x +
b
sin
x = c |
| To
solve the trigonometric equations which are linear in sin
x and cos
x, and where, a,
b,
and c
are real |
| numbers
we can use the two methods, |
| a)
introducing an auxiliary angle, and b)
introducing new unknown. |
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| Introducing new unknown
t
= tan x/2
example |
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Example: Solve the equation,
5 sin x
-
4 cos x = 3.
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Solution: Given equation is of the form
a
cos x + b sin x = c
therefore parameters are, a
= -
4, b =
5 and c
= 3, after introducing new unknown
tan x/2 = t
and substituting the values of the parameters into equation
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| (a +
c) ·
t2
-
2b · t + (c -
a) = 0
gives (-
4 + 3) ·
t2
-
2 · 5 · t + [3 - (-
4)] = 0 |
| or
t2
+ 10t -
7 = 0,
t1,2
= -
5 +
Ö25
+ 7 = -
5 ± 4Ö2. |
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Obtained values for variable
t
we plug into substitutions,
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tan x/2 = t1,
x/2 = tan-1
(t1)
or x =
2arctan(-
5 -
4Ö2) |
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x = 2 · (-
84°38′21″ + k · 180°) =
-
169°16′42″ + k · 360°,
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tan x/2 = t2,
x/2 = tan-1
(t2)
or x′ =
2arctan(-
5 + 4Ö2) |
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x′ =
2 · (33°17′56″ + k · 180°) =
66°35′53″ + k · 360°. |
| The same result we obtain using the method of introducing the auxiliary angle
j. Plug the given parameters
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| a
= -
4, b =
5 and c
= 3 into tan
j
= b/a,
tan j =
5/(-
4) => j =
-51°20′24″, cos j =
0.624695, |
| then
from the equation |
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cos
(x + 51°20′24″) =
(3/-
4) · 0.624695 |
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or x
+ 51°20′24″ =
arccos [(3/-
4) · 0.624695], |
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| thus,
x +
51°20′24″ =
+
117°56′18″ + k · 360° => x =
+
117°56′18″
-
51°20′24″ + k · 360°. |
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Trigonometric equations of the form a
cos x + b sin x = c
we do not solve using the identity
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since that way given equation becomes quadratic with four solutions but only two of
them satisfy it. |
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We will solve the equation from the previous example using this method anyway. |
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Example: Solve the equation,
5 sin x
-
4 cos x = 3
by substituting |
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Solution: Squaring both sides of an equation may introduce extraneous
or redundant (not needed) solutions.
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| By plugging the results into given equation show that only solutions
b) and c) satisfy the equation what match with previous results obtained using another two methods.
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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 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(-1/4)
, x =
-115°
+ tan-1
(-1/4) =
- 129°2′10″ + k · 180°. |
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Example: Find the solution
of the equation, 2
sin
(x
+ 60°) · cos
x = 1.
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Solution: Applying products as sums formula
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| 2 · (1/2)
[sin
(x
+ 60°
+ x)
+ sin
(x
+ 60° - x)] = 1
or sin
(2x
+ 60°)
+ sin 60° = 1
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| then
sin
(2x
+ 60°) = 1
- Ö3/2,
2x
+ 60° =
sin-1(1
- Ö3/2) + k · 360° |
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and 2x′
+ 60° =
180°
- sin-1(1
- Ö3/2) + k · 360° |
| so,
the solution is 2x =
-
60°
+ sin-1(1
- Ö3/2) + k · 360°,
x =
-
26°9′1″ + k · 180°, |
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2x′ =
120° - sin-1(1
- Ö3/2) + k · 360°,
x′ = 56°9′1″ + k · 180°,
k Î Z. |
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Example: Find the solution
of the equation, |
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| Solution: Using
identities |
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| given
equation becomes 2 · (1
+ cos x) - Ö3 · cot x/2 = 0, |
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| therefore,
1
+ cos x
= 0,
cos
x
= -
1,
x
= p
+ k · 2p, |
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and 2sin
x -
Ö3
= 0,
sin
x
= Ö3/2,
x
= p/3
+ k · 2p
and x′ = 2p/3
+ k · 2p,
k Î Z. |
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| Trigonometry
contents B |
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