==> (from the graph)

==>

Let

(1)

Take the cosine of both side

Solution:

e) No solution at all

d) A solution x
(0, 1]

c) A solution x

b) Two real solutions

a) Exactly one solution which is a negative integer

The number of solutions of the equation

6.6.2 (001-T2-11
$6.6)

Square both sides

Move to the other side

Therefore, equation (1) becomes

==>

==>

==>

Similarly, let

==>

6.6.1 (001-T2-7
$6.6)

Section
6.6

==>

Recall:

Solution:

6.5.3

Mathcad check:

Expand the cosine of the difference

Apply cosine to both sides.

Rewrite the eqn.

Therefore, the sum is

==>

or

or

==>

==>

Solution:

e)

d)

c)

b)

a)

The sum of all the solutions of the equation
in the interval
is equal to

Rewrite the eqn

==>

The zeros of f(x) are obtained by setting f(x) = 0 and
solving for x.

Solution:

Find the zeros of the function
in the interval [0, 2p)

6.6.5

Note:
is not a solution since
is not defined.

or

==>

or

==>

Solve for x.

==>

k is an integer.

==>

Solution:

Find the set of solutions of the equation sin3x = 1.

6.6.6

and

==>

==>

or

or

Solution:

Find the number of solutions of the equation
over the interval [0, 3p
/2)

6.6.3

If we substitute x = -1 and x =
in equation (2), we find that only x = -1 satisfies
the equation.

or

or

==>

==>

==>

Divide by sin(x) and consider that sin(x) =0 may include
solutions.

==>

==>

Solution:

Find all the solutions of the equation
,

6.6.4

The number of solutions is 4.

or

==>

or

==>

c)

b)

a)

tan 15o
=

6.3.2 (001-T2-4
$6.3)

Figure 1

The cosine is positive and the sine is negative in quadrant
IV. Thus from the graph

Solution:

e)

d)

c)

b)

a)

If
,
,
, then
is equal
to:

6.3.1 (001-T2-3
$6.3)

where the choice of (+) or (-) depends on the quadrant
in which the half-angle
terminates.

Solution:

e)

d)

c)

b)

a)

6.3.3 (001-T2-6
$6.3)

Mcad check:

Rationalize the denominator

Solution:

e)

d)

c)

b)

a)

6.2.1 (001-T2-9
$6.2)

Section
6.2

Solution:

e)

d)

c)

b)

a)

The expression

6.1.1 (001-T2-18
$6.1)

Section
6.1

A. M. Farhat

Solution of Old Exams Problems

Chapter
6

Math002

Section
6.3

Therefore, by substituting (1) and (2) we have

(2)

and

Let

Working on the second term.

(1)

Working on the second term.

Solution:

e)

d)

Thus,

The sine and cosine of a
are
negative ==> a
is in quadrant III. From the values of the sine and
cosine we compute the reference angle of a. The reference angle is
.

Therefore,

Solution:

e) k = 2, a
=

d) k = 2, a
=

c) k = 2, a
=

b) k = 2, a
=

a) k = 2, a
=

When writing
in the form
of
,
, then

6.4.1 (001-T2-1
$6.4)

Section
6.4

Solve the equation
for x.

6.5.2

Thus,

if

(Only if x is in the range of arcsine)

if
.

Recall:

Solution:

e)

d)

c)

b)

a)

6.5.1 (001-T2-5
$6.5)

Section
6.5

e) 2

d) 6

c) -6

b) 3

a) -3

The maximum value of the function
is:

6.3.5 (001-T2-17
$6.3)

Solution:

e)

d)

c)

b)

a)

The exact value of
is:

6.3.4 (001-T2-16
$6.3)

Thus,

==>

Graph of

Graph of 3sin(4x)

==>

==>

The maximum values will occur when sin (4x) = 1

Solution:

Find all the x values at which the function has maximum
values

6.3.5-1 (001-T2-17
$6.3)

Thus, the maximum value of f(x) is 3.

Use the double angel identity of the sine

Use the cofunction identity.

Solution: