Transcript Sec 4.4
Chapter 4
Analytic Trigonometry
Section 4.4
Trigonometric Equations
Basic Trigonometric Equations
When you first learn to solve equations you learn how to solve simple linear
equations of the form:
ax + b = c
x = (c – b)/a
has solution
Which was solved by a process we call isolating the variable. Later on you learn
factoring as a way to change a more complicated equation that has square terms
(or higher ) to a series of simple linear equations and solve them.
We will do the same thing with trigonometric equations. We will learn how to solve
the basic form of a trigonometric equation then move up to more complicated
versions. The first type of trigonometric equations we will learn to solve are of the
form:
sin x = c
or
cos x = c
or
tan x = c
These can be solved by hand for certain values of c and by a calculator for all
values of c.
In general equations of this type have either no solutions or an infinite number.
The idea will be to find all solutions between 0 and 2 (i.e. 0 and 360 if using
degrees). Once those solutions are found all other solutions can be expressed as
the solution you found plus a multiple of 2 (360 if using degrees).
The equations: sin x = c and cos x = c
can be solved by hand for the values of
c being:
1
2
3
0, 1, ,
,
2
2
2
The example to the right is how to
solve:
3
2
Solutions:
x
5
2k
6
and
x
1. Draw unit circle.
2. Draw horizontal or
vertical line the correct
distance on x or y axis.
5
6
1
2
3. Find angles where
line hits the unit circle.
1
2
3
cos x
2
c
cos x 2 3
7
2k
6
4. Add 2k to each
angle.
7
6
x 150
If the value c is not one of the numbers
above we use the calculator and the
inverse trigonometric functions. Below 157.976
and to the right we show how to solve:
k 360 and x 210 k 360
2. Draw the angle.
3. Draw horizontal or
22.0243 vertical line the correct
distance on x or y axis.
3
8
sin x = ⅜ (i.e. c = ⅜)
1.Compute x=sin-1(⅜) .384397 (radians)
22.024
4. Find the other angle
where line hits unit
circle.
180-22.024=157.976
- .384397 = 2.7572
Solutions: x 22.0243 k 360 and x 157.976 k 360 x .384397 2k and x 2.7572 2k
Tangent is
negative in
Quadrants
II and IV
Tangent is positive in the
Quadrants I and III
Take Reciprocals
These are
called two
specific
solutions.
For equations
involving the other
trigonometric
functions we can
take reciprocals
and apply the
reciprocal identities
to solve them.