Answers and Explanations 185
Answers
101–200
166.
x = π – 0.322, 2π – 0.322, 3
4
π , 7
4
π
To solve 3 sec
2 
x + 4 tan x = 2 for x over the interval [0, 2π], begin by using the identity
sec
2 
x = 1 + tan
2 
x and make one side of the equation equal to zero:
3 1
4
2
3 3
4
2
3
4
1 0
2
2
2
+
(
) +
=
+
+
=
+
+ =
tan
t an
tan
t an
tan
t an
x
x
x
x
x
x
Next, use the quadratic formula to find tan x:
tan
( )( )
( )
( )
( )
x =
− ±
−
= − ±
= − ±
4
4 4 3 1
2 3
4
4
2 3
4 2
2 3
2
Therefore, tan x = –1 and − 1
3
.
The solutions to the equation tan x = –1 are x = 3
4
π and x = 7
4
π . To solve tan x = − 1
3
,
take the inverse tangent of both sides:
tan
tan
.
x
x
= −
=
−
( )
≈ −
−
1
3
1
3
0 322
1
Note that this solution isn’t in the given interval. The solutions that are in the given
interval and belong to Quadrants II and IV (where the tangent function is negative) are
x = π – 0.322 and x = 2π – 0.322.