6. In examining the equation sin(2
x
) =
√
3 cos(
x
), not only do we have different circular func
tions involved, namely sine and cosine, we also have different arguments to contend with,
namely 2
x
and
x
. Using the identity sin(2
x
) = 2 sin(
x
) cos(
x
) makes all of the arguments the
same and we proceed as we would solving any nonlinear equation – gather all of the nonzero
terms on one side of the equation and factor.
sin(2
x
)
=
√
3 cos(
x
)
2 sin(
x
) cos(
x
)
=
√
3 cos(
x
)
(Since sin(2
x
) = 2 sin(
x
) cos(
x
).)
2 sin(
x
) cos(
x
)

√
3 cos(
x
)
=
0
cos(
x
)(2 sin(
x
)

√
3)
=
0
from which we get cos(
x
) = 0 or sin(
x
) =
√
3
2
. From cos(
x
) = 0, we obtain
x
=
π
2
+
πk
for
integers
k
. From sin(
x
) =
√
3
2
, we get
x
=
π
3
+2
πk
or
x
=
2
π
3
+2
πk
for integers
k
. The answers
7
As always, experience is the greatest teacher here!
8
As always, when in doubt, write it out!
10.7 Trigonometric Equations and Inequalities
865
which lie in [0
,
2
π
) are
x
=
π
2
,
3
π
2
,
π
3
and
2
π
3
. We graph
y
= sin(2
x
) and
y
=
√
3 cos(
x
) and,
after some careful zooming, verify our answers.
y
= cos(3
x
) and
y
= cos(5
x
)
y
= sin(2
x
) and
y
=
√
3 cos(
x
)
7. Unlike the previous problem, there seems to be no quick way to get the circular functions or
their arguments to match in the equation sin(
x
) cos
(
x
2
)
+ cos(
x
) sin
(
x
2
)
= 1. If we stare at
it long enough, however, we realize that the left hand side is the expanded form of the sum
formula for sin
(
x
+
x
2
)
. Hence, our original equation is equivalent to sin
(
3
2
x
)
= 1. Solving,
we find
x
=
π
3
+
4
π
3
k
for integers
k
. Two of these solutions lie in [0
,
2
π
):
x
=
π
3
and
x
=
5
π
3
.
Graphing
y
= sin(
x
) cos
(
x
2
)
+ cos(
x
) sin
(
x
2
)
and
y
= 1 validates our solutions.
8. With the absence of double angles or squares, there doesn’t seem to be much we can do.
However, since the arguments of the cosine and sine are the same, we can rewrite the left
hand side of this equation as a sinusoid.
9
To fit
f
(
x
) = cos(
x
)

√
3 sin(
x
) to the form
A
sin(
ωt
+
φ
) +
B
, we use what we learned in Example
10.5.3
and find
A
= 2,
B
= 0,
ω
= 1
and
φ
=
5
π
6
. Hence, we can rewrite the equation cos(
x
)

√
3 sin(
x
) = 2 as 2 sin
(
x
+
5
π
6
)
= 2,
or sin
(
x
+
5
π
6
)
= 1.
Solving the latter, we get
x
=

π
3
+ 2
πk
for integers
k
.
Only one of
these solutions,
x
=
5
π
3
, which corresponds to
k
= 1, lies in [0
,
2
π
).
Geometrically, we see
that
y
= cos(
x
)

√
3 sin(
x
) and
y
= 2 intersect just once, supporting our answer.
y
= sin(
x
) cos
(
x
2
)
+ cos(
x
) sin
(
x
2
)
and
y
= 1
y
= cos(
x
)

√
3 sin(
x
) and
y
= 2
We repeat here the advice given when solving systems of nonlinear equations in section
8.7
– when
it comes to solving equations involving the trigonometric functions, it helps to just try something.
9
We are essentially ‘undoing’ the sum / difference formula for cosine or sine, depending on which form we use, so
this problem is actually closely related to the previous one!
866
Foundations of Trigonometry
Next, we focus on solving inequalities involving the trigonometric functions. Since these functions
are continuous on their domains, we may use the sign diagram technique we’ve used in the past to
solve the inequalities.