Answers and Explanations 183
Answers
101–200
Likewise, there’s a solution to the equation sin( )
y = 3
5
in the second quadrant because
the function also has positive values there, namely π – 0.644; adding 2π gives you the
other solution, 3π – 0.644. Therefore, y = 0.644, π – 0.644, 2π + 0.644, and 3π – 0.644 are
all solutions.
Last, substitute 2x into the equations and divide to get the solutions for x:
2
0644
0 322
x
x
=
=
.
.
2
0644
2
0 322
x
x
= −
= −
π
π
.
.
2
2
0 644
0 322
x
x
=
+
= +
π
π
.
.
2
3
0 644
3
2
0 322
x
x
=
−
=
−
π
π
.
.
The solutions are x = 0.322,
π
2
0 322
− .
, π + 0.322, and 3
2
0 322
π − .
.
164.
x = 0.321, 2
3
0 321
π − .
, 2
3
0 321
π + .
, 4
3
0 321
π − .
, 4
3
0 321
π + .
, 2π – 0.321
To solve 7 cos(3x) – 1 = 3 for x over the interval [0, 2π], first isolate the term involving
cosine:
7
3
1 3
3
4
7
cos( )
cos( )
x
x
− =
=
You can also use the substitution y = 3x to simplify the equation. Because 0 ≤ x ≤ 2π, it
follows that 0 ≤ 3x ≤ 6π so that 0 ≤ y ≤ 6π. Use the substitution and take the inverse
cosine of both sides:
cos
cos
.
y
y
=
=
( )
≈
−
4
7
4
7
0 963
1
It follows that in the interval [0, 6π], y = 2π + 0.963 and y = 4π + 0.963 are also solutions
because adding multiples of 2π makes the resulting angles fall at the same places on
the unit circle.
Likewise, there’s a solution to the equation cos y = 4
7
in the fourth quadrant
because cosine also has positive values there, namely y = 2π – 0.963. Because y = 2π –
0.963 is a solution, it follows that y = 4π – 0.963 and y = 6π – 0.963 are also solutions.
Therefore, you have y = 0.963, 2π – 0.963, 2π + 0.963, 4π – 0.963, 4π + 0.963, and
6π – 0.963 as solutions to cos y = 4
7
. Last, substitute 3x into the equations and divide
to solve for x: