College Trigonometry, 2011a

10.7 Trigonometric Equations and Inequalities 871
arcsin(2x) = π 3
sin (arcsin(2x)) = sin ( )
π
2x =
x =
√
3
3
2
Since sin(arcsin(u)) = u
√
3
4
Graphing y = arcsin(2x) and the horizontal line y = π √
3 , we see they intersect at 3
4 ≈ 0.4430.
2. Our first step in solving 4 arccos(x) − 3π = 0 is to isolate the arccosine. Doing so, we get
arccos(x) = 3π 4 .Since 3π 4
is in the range of arccosine, we may apply Theorem 10.26
arccos(x) = 3π 4
cos (arccos(x)) = cos ( )
3π
4
√
x = − 2
2
Since cos(arccos(u)) = u
√
The calculator confirms y = 4 arccos(x) − 3π crosses y = 0 (the x-axis) at − 2
2 ≈−0.7071.
y = arcsin(2x) andy = π 3
y = 4 arccos(x) − 3π
3. From 3 arcsec(2x − 1) + π =2π, we get arcsec(2x − 1) = π 3
. As we saw in Section 10.6,
there are two possible ranges for the arcsecant function. Fortunately, both ranges contain π 3 .
Applying Theorem 10.28 / 10.29, weget
arcsec(2x − 1) = π 3
sec(arcsec(2x − 1)) = sec ( )
π
3
2x − 1 = 2 Since sec(arcsec(u)) = u
x = 3 2
To check using our calculator, we need to graph y = 3 arcsec(2x − 1) + π. Todoso,wemake
use of the identity arcsec(u) = arccos ( 1
( )
u)
from Theorems 10.28 and 10.29. 16 We see the graph
of y = 3 arccos 1
2x−1
+ π and the horizontal line y =2π intersect at 3 2 =1.5.
16 Since we are checking for solutions where arcsecant is positive, we know u =2x − 1 ≥ 1, and so the identity
applies in both cases.