Similar Triangles, Right Triangles, And The Definition Of The Sine, Cosine And Tangent Functions Of Angles Of A Right Triangle Worksheets With Answer Key Page 36
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43The technique for solving these equations is to transform them into the form of the simpler equations
just considered by making a change of variables.
2
θ
θ
π
=
≤
<
Example: Find all solutions to
sin
2
in the interval
0
2
.
2
φ
θ
Solution: Make the change of variables
= 2
so the equation becomes
2
φ
sin =
.
2
The general solution to the original equation is
π
π
3
φ
π
φ
π
=
+
=
+
2
n
,
2
n
,
n an integer.
4
4
Thus, the general solution to the original equation is
π
π
3
θ
π
θ
π
=
+
=
+
2
2
n
,
2
2
n
,
n an integer.
4
4
or
π
π
3
θ
π
θ
π
=
+
=
+
n an integer.
n
,
n
,
8
8
θ
π
≤
<
The solutions in the interval
are
0
2
π
π
π
π
3
9
11
θ
=
,
,
,
.
8
8
8
8
Problems:
θ
π
≤
<
In problems 192-195, find all solutions in the interval
0
2
.
1
θ
=
θ
=
−
193.
tan
3
0
192.
cos
2
2
θ
1
3
=
−
θ
=
−
194.
sin
195.
sin
4
2
2
2
We now extend our techniques to solve equations that yield to rearranging and factoring.
θ
θ
θ
θ
π
≤
<
Example: Find all solutions of cos
tan
= cos
in
0
2
.
θ
θ
θ
Solution: Rewrite cos
tan
= cos
as
θ
θ
θ
cos
tan
- cos
= 0
and factor to obtain
θ
θ
cos
(tan
- 1) = 0.
θ
θ
Either cos
= 0 or tan
- 1 = 0.
We solve these equations separately in the stated interval.
π
π
3
θ
θ
cos =
=
0
gives solutions
,
.
2
2
π
π
5
θ
θ
−
=
=
tan
1
0
(or
tan
) 1
gives
solutions
and
.
4
4
θ
π
≤
<
Thus, the solutions in the interval
0
2
are
π
π
π
π
3
5
θ
θ
=
=
,
and
,
.
2
2
4
4
- 36 -
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