==> (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
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)
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: