Part II: The Answers
176
Answers
101–200
gives you 2 cos x + 1 = 0 so that cos x = − 1
2
, which has the solutions x = 2
3
π and
4
3
π . Setting the second factor equal to zero gives you cos x + 1 = 0 so that
cos x = –1, which has the solution x = π.
143.
π
4
, 7
12
π , 11
12
π , 5
4
π , 19
12
π , 23
12
π
Begin by making the substitution y = 3x. Because 0 ≤ x ≤ 2π, you have 0 ≤ 3x ≤ 6π so that
0 ≤ y ≤ 6π. Finding all solutions of the equation tan y = –1 in the interval [0, 6π] gives
you y = 3
4
π , 7
4
π , 11
4
π , 15
4
π , 19
4
π , and 23
4
π .
Next, take each solution, set it equal to 3x, and solve for x: 3
3
4
x = π so that x = π
4
;
3
7
4
x = π so that x = 7
12
π ; 3
11
4
x = π so that x = 11
12
π ; 3
15
4
x = π so that x = 5
4
π ;
3
19
4
x = π so that x = 19
12
π ; and 3
23
4
x = π so that x = 23
12
π .
144.
π
4
, 3
4
π , 5
4
π , 7
4
π
Begin by making one side of the equation equal to zero and then factor:
cos
cot( )
cos
cot
cos
cos
sin
cos
sin
2
2
2
2
0
2
2
2
0
2 1
1
2
x
x
x
x
x
x
x
x
( )=
−
=
−
=
−
x x
(
) = 0
Next, make the substitution y = 2x:
cos
sin
y
y
1
1
0
−





 =
Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 2x ≤ 4π so that 0 ≤ y ≤ 4π.
Setting the first factor equal to zero gives you cos y = 0, which has the solutions y = π
2
,
3
2
π , 5
2
π , and 7
2
π . Take each solution, set it equal to 2x, and solve for x: 2
2
x = π so that
x = π
4
; 2
3
2
x = π so that x = 3
4
π ; 2
5
2
x = π so that x = 5
4
π ; and 2
7
2
x = π so that x = 7
4
π .
Proceed in a similar manner for the second factor: 1
1
0
−
=
sin y
so that sin y = 1, which
has the solutions y = π
2
and 5
2
π . Now take each solution, set it equal to 2x, and solve for
x: 2
2
x = π so that x = π
4
, and 2
5
2
x = π so that x = 5
4
π .