Math 1330
Section 6.3
Solving Trig Equations
This section concerns solving equations with trig functions. Note that these equations are
not identities, just equations in the usual sense. The basic equation you need to
understand how to solve is one of the form:
Trig function (x) = number
Solving for x means finding the number (angle in degrees or radians) whose trig function
is that number. The process of solving involves two steps which we will illustrate by
solving
1
cos( )
2
x
= −
.
Step One: Identify the domain for one cycle of the trig function. In this case that is the
interval
[0,2 ]
π
.
Then find the values of x in this interval for which cosine has the value –1/2.
For this case, we know from the table of values for cosine and the graph that
cosine takes on the value of –1/2 at
2
3
x
π
=
and
4
3
x
π
=
.
Step Two: Since the trig functions are periodic, there are corresponding solutions to this
equation in each period of the function. So, every angle that is one of the solution angles
in the basic cycle plus an integer multiple of the period of the function is also a solution.
In simple terms, since the period for cosine is
2
π
, the general solutions are
2
2
3
x k
π
π
= +
and 4
2
3
x k
π
π
= +
for all integers, k.
We can also solve trig equations when the numbers are not as “pretty”. Note that in the
above example, the number –1/2 corresponds to the cosine value of a special angle. To
solve the equation when the number does not correspond to the trig function value of a
special angle, we use the inverse functions and a calculator to solve the equation.
Note that the homework requires solving some of the equations where the angle is not a
special angle, and a calculator is needed. The test and final will both use special angles so
no calculator will be needed.
Let me illustrate how to solve these types of equations by solving
sin( ) 0.8
x
=
using a
calculator and the inverse sine function.
First step is to state the range of the inverse function in question. In this case the range of
(
)
1
sin
x
−
is
[ , ]
2 2
π π
−
.
