Solving Trig Equations
Section 1
In algebra class you spent a lot of time learning to solve equations.
For example
2x + 1 = 15
which has the solution x = 7, and x
2
+1 = 10 which has two solutions x = 3
and x = 3.
Keep in mind that we are finding values of x that will satisfy the equation.
In
this unit we will learn to solve equations where the unknown is “inside” a trig function.
For example we will learn to solve sin(2x + 1) = 0.
The key to solving equations like this
is to use the graph of sine or cosine to get started.
The figure to the right shows the graph of y = sin(x).
When we solve the equation sin(x) = 0 we are finding
values of x for which sin(x) is zero.
On the graph this
means values of x where the ycoordinate is zero.
We
don’t have to do any work.
All we do is read the x
values from the graph.
Therefore six solutions to
sin(x) = 0 are x = 2
π
, –
π
, 0 ,
π
, 2
π
, 3
π
and we could
continue this pattern to get as many solutions as
desired.
Suppose we want to solve the equation cos(x) = 1.
We would look at the graph of y=cos(x) and read the
x values that have a y coordinate equal to 1.
So the
four solutions are x = 2
π
, 0, 2
π
, 4
π
and infinitely
many more following the same pattern.
Exercises
:
1.
Give three solutions to sin(x) = 1.
2.
Give three solutions to cos(x) = 0.
3.
Give five solutions to sin(x) = 1.
4.
Give four solutions to cos(x) = 1.
Most of the trig equations we will encounter in applications are not quite as simple as
those shown above.
But solving those will be the basis for solving more complicated
equations.
For example suppose we want to solve the equation sin(2x – 1) = 0.
The key
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here is to recognize that the equation has the form sine of “something” equals 0, or in
symbols
sin(
θ
) = 0.
We can use the graph of sine to solve sin(
θ
) = 0 and get
θ
= 0
or
θ
=
π
or
θ
= 2
π
or
θ
= 3
π
etc.
In the original equation sin(2x – 1) = 0 we have
θ
= 2x – 1. Therefore
from the line
above we can write
2x – 1 = 0
2x – 1 =
π
2x – 1 = 2
π
2x –1 = 3
π
which we can solve to get
2x = 1
2x =
π
+ 1
2x = 2
π
+ 1
2x = 3
π
+ 1
x =
1
2
x =
π
+
1
2
x =
2
π
+
1
2
x =
3
π
+
1
2
We believe that these are solutions to sin(2x – 1) = 0.
Recall from algebra that you can
check a solution by substituting the value back into the equation and see if the equation is
satisfied.
For example to check the solution x =
1
2
above, we would substitute
1
2
for x in
on the left side of the equation sin(2x – 1) = 0 and we get
sin(2(
1
2
)
−
1)
=
sin(1
−
1)
=
sin(0)
=
0
which means that x =
1
2
is a solution.
To check
x =
π
+
1
2
we would calculate
sin(2(
π
+
1
2
)
−
1)
=
sin((
π
+
1)
−
1)
=
sin(
π
)
=
0
.
Another example
:
Find three positive solutions to cos(
π
x) = 1.
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 Fall '08
 RUSSEL
 Algebra, Trigonometry, Equations, Sin, Inverse function, Inverse trigonometric functions, θ

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