
Trigonometry 



Trigonometric
Equations 
Basic Trigonometric
Equations 
The
equation cos
x = a 






Trigonometric
equations

An equation that involves one or more trigonometric functions, of an unknown arc, angle or number, is called trigonometric equation.

Basic trigonometric
equations

The equation cos
x = a, 1
<
a <
1

The solutions of the equation are arcs
x
whose function's value
of cosine equals a.

Infinite many arcs whose cosine value equals a end in points,
P 
and P′, that are 

x
=
a^{rad}
+ k
·
2p
and x′
=  a^{rad}
+ k
·
2p,
k Î
Z. 


This is the set of the general solutions of the given equation. 
For
k = 0 follows the basic solutions of the equation 

x_{0}
=
a^{rad}
and x_{0}′
=  a^{rad}.






Therefore, if cos
x = a, 1
< a
<
1 then,
x = ± a^{rad}
+ k ·
2p
= + arccos a, k Î
Z.

For example if, a =
1,
then, cos
x = 1
=> x = p
^{}
+ k
·
2p,
k Î
Z,

a = 0
cos x =
0 =>
x = p/2
+ k
·
p,
k Î
Z,

or a =
1
cos x =
1 =>
x = k
·
2p,
k Î
Z.


Since cosine function passes through all values from range
1
to 1
while arc x
increases from 0
to p, one of the arcs from this interval must satisfy the equation
cos x = a.

This arc, denoted x_{0},
we call the
basic solution.

Thus, the basic solution of the equation
cos x = a,
1
<
a
<
1
is the value of inverse cosine function,

x_{0}
=
arccos a
or x_{0}
= cos^{1}
a,

that is, an arc or angle (whose cosine equals
a) between
0
and p
which is called the principal
value.

Scientific calculators are equipped with the
arccos (or
cos^{1}) function which, for a given argument between
1
and 1, outputs arc (in radians) or angle (in degrees) from the range
x_{0} Î
[0,
p].


Example: Solve
the equation, cos
x = 
0.5.

Solution: In the unit circle in
the below figure shown are the two arcs, of which cosine value equals

0.5, that represent the basic solutions of the given equation

x_{0}
= 120°
or x_{0}′ =
120°

while the abscissas of the intersection points of the line
y = 
0.5 with the graph of cosine function represent
the set of the general solution

x
= +
120°
+ k
·
360°
or x
= +
2p/3
+ k
·
2p,
k Î
Z.


The same results we obtain by using calculator if we set DEG then input


0.5
INV cos
(or cos^{1})
=> x_{0}
= 120°
and x_{0}′ =
120° that
are the basic solutions.

Or we input the same while calculator is set in RAD mode to get the arc in radians that is

x_{0}
= 2.094395102^{rad}
= 2p/3^{rad}.









Precalculus
contents G 



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