Download 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.