CHAPTER 5
LESSON 1
Teachers Guide
Solving Trigonometric Equations Using Graphing Technology
AW 5.1
MP 5.1
Objective:
 To solve trigonometric equations graphically over a domain of length 2  , and
where the domain is the set of all real numbers.
Definitions
There are two kinds of trigonometric equations:
1. Identities: Identities are equations that are valid for all values of the variable for
which the equation is defined; for example,
1
sin2θ = 2sinθcosθ and cscθ =
sinθ
2. Conditional Equations: Conditional Equations are valid only for certain values
3
of the variable; for example, cosx =
. Solutions for these equations can be found
2
using techniques similar to algebraic equations.
Prerequisite
Use a graphing calculator to solve the algebraic equation 2 x 2  x  3  0 .
The x-intercepts, ‘roots’, or ‘zeros’ of the
graph are the solutions to the original
equation;
x=
3
2
or -1
Trigonometric
equations can also be solved graphically.
Investigate
a) Solve the equation sin x  0.5 , 0  x  2 , graphically, in 2 ways. Express the
answer accurate to 2 decimal places.
Method 1 (Intersection Method)
First draw the graph of y  sin x and
y  0.5
Method II (x-intercept method)
First rewrite sin x  0.5 as
y  sin x  0.5 and graph
Y1 = sinx
Y2 = -0.5
To set the window:
Y1 = sinx + 0.5

Use the given domain for x-min and xmax values.

π 1
π
Recall that sin = so x - scl = is a
6 2
6

good choice.
Amplitude of y = sinx is 1 so
y - min = -2 , y - max = 2 .
Note: Experiment with your window until
the graph is clearly visible. Don’t assume
intersection or other properties of your
graph
Where are the ‘solutions’ to the equation
sinx = -0.5 using this method?
Solution
:The solutions are the x-coordinates of the
intersection points. Use the ‘intersect’
option on the ‘Calculate’ menu of your
calculator.
X 3.67 or X 5.76
Using this method the x-intercepts are
the solutions.
X
3.67 or X
5.76
Method I or II may be used.
b) Express the solutions in exact values
sin
π 1
=
6 2
Sine is negative in Quadrants 3 and 4
π 7π
=
6
6
π 11π
Or x = 2π - =
6
6
So x = π +
c) Change the domain to 4  x  4 , and graph y  sin x  0.5 again. What do you
notice?
The graph is periodic (it repeats). There are many more solutions, 8 instead of 2. The new
solutions differ from the original solutions,
7π
6
and
11π
6
by multiples of 2 , the
period of the graph; for example, one new solution is 7π + 2π =
19π
.
6
d) What if there were no restrictions on the domain? How many solutions would there
be for sin x  0.5 ?
An infinite number of solutions
e) How could all these solutions be written?
x=
7π
+ 2nπ or
6
11π
+ 2nπ , where n is any integer.
6
Definition
The solution of a trigonometric equation over all real numbers is called the general
solution.
List the steps used to find the general solution to a trigonometric equation using the
graphic calculator?



Set the window to graph at least 1 full cycle
Find the solutions in this 1st cycle
All other solutions will differ from these by multiples of the period.
Example 1:
Solve graphically for x, to 3 decimal places: cos3 x  3cos x  1  0
Technology tips
1. Enter powers such as sin 2 x as (sin(x))2 on the calculator.
2. Most calculators provide a window setting that automatically scales the x-axis in terms
of  . This option can provide a quick first look at the graph if a specific domain is not
stated. The window can then be adjusted if desired . See ZTRIG in the ZOOM menu
(TI), E TRIG (Sharp9600); F3, F2 from GRAPH (Casio), and Trig from VIEW (HP38).
The period is 2π . This question requires a general solution without exact values. The 2
solutions for 0  x  2π are x = 1.216 and x = 5.067 . The general solution is
x = 1.216 + 2nπ or x = 5.067 + 2nπ , where n is any integer. Y1 = cos 3 x - 3cosx + 1
Example 2:
Solve for x graphically, in exact values, with 0  x  2 if tan 2 x  3 .
Y1 = tan 2 x - 3
The first two solutions, as decimals, are :
x = 1.0471976 or x = 2.0943951
The x-scale for the graph was
π
π
2π
and the graph appears to have roots at
and
. The
3
6
3
decimal solutions correspond to these exact values..
The period of the graph is  .
The 4 solutions are
x=
π 2π π
2π
+π
,
, + π , or
3 3 3
3
x=
π 2π 4π
5π
,
,
, or
3 3
3
3