Which of the following trigonometric inequalities has no solution over the interval 0 ≤ x ≤ 2pi radians?
A. Cos(x)<1
B. cos(x/2)<1
C. csc(x)<1
D. csc(x/2)<1

A .cos(x)<1

Step-by-step explanation:

According to the first inequality

cos(x)<1

x < arccos 1

x<0

This therefore does not have a solution within the range 0 ≤ x ≤ 2pi

x cannot be leas than 0. According to the range not value, 0≤x which is equivalent to x≥0. Thus means otvis either x = 0 or x> 0.

For the second option

.cos(x/2)<1

x/2< arccos1

x/2<0

x<0

This inequality also has solution within the range 0 ≤ x ≤ 2pi since 0 falls within the range of values.

For the inequality csc(x)<1

1/sin(x) < 1

1< sin(x)

sinx>1

x>arcsin1

x>90°

x>π/2

This inequality also has solution within the range 0 ≤ x ≤ 2pi since π/2 falls within the range of values

For the inequality csc(x/2)<1

1/sin(x/2) < 1

1< sin(x/2)

sin(x/2)> 1

x/2 > arcsin1

X/2 > 90°

x>180°

x>π

This value of x also has a solution within the range.

Therefore option A is the only inequality that does not have a solution with the range.

8

step-by-step explanation:

just put √(2) * √(32) into a calculator. that will make the answer be 8.