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 37
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43θ
θ
θ
π
≤
<
2
Example: Find all solutions of 2sin
+ 1 = 3sin
in
.
0
2
Solution: Rewrite the equation as
θ
θ
2
2sin
- 3sin
+ 1 = 0
θ
and factor as a quadratic in sin
to obtain
θ
θ
(2sin
- 1)(sin
- 1) = 0.
Set each factor equal to 0 and solve the two equations
θ
θ
2sin
- 1 = 0 and sin
- 1 = 0
π
π
5
θ
π
≤
<
in the interval
0
2
. The first yields solutions
and
, while the second
6
6
π
θ
θ
π
≤
<
yields
=
. The solutions in the interval
0
2
are
2
π
π
π
5
θ =
and θ =
,
.
6
6
2
Problems:
θ
π
≤
<
In problems 196-200, find all solutions in the interval
0
2
θ
θ
θ
2
2
196. 4sin
= 1
197. 2cos
- cos
= 1
θ
θ
θ
θ
θ
θ
4
2
198. 4cos
= 3 - cos
199. sin
cos
+ cos
= 1 + sin
θ
θ
2
200. sin
2
= sin 2
There are two situations in which identities are commonly used to solve trigonometric equations.
1. When the equation involves more than one trigonometric function, an identity may be used to
rewrite the equation in terms of just one function.
2. When the trigonometric equation involves an unknown angle θ and its multiples, we may use
the double-angle, half-angle and addition formulas to rewrite the equation in terms of just one
angle.
These techniques are illustrated in the following examples.
θ
θ
θ
π
≤
<
2
Example: Find all solutions of sec
+ tan
= 1 in
0
2
.
θ
θ
2
2
Solution: We use the identity sec
= 1 + tan
to rewrite the equation as
θ
θ
2
1 + tan
+ tan
= 1.
Subtract 1 from both sides and factor to get
θ
θ
tan
(tan
+ 1) = 0.
θ
θ
θ
π
≤
<
Thus tan
= 0 or tan
+ 1 = 0, so the solutions in
are
0
2
π
π
3
7
θ
π
θ
=
=
, 0
and
,
.
4
4
θ
θ
θ
π
≤
<
Example: Find all solutions of sin
- cos2
= 0 in the interval
0
2
.
θ
θ
2
Solution: We use the identity cos2
= 1 - 2 sin
to rewrite the equation as
θ
θ
2
sin
- (1 - 2 sin
) = 0
θ
θ
which is now in terms of
(and not 2
). Solving by factoring,
- 37 -
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