Question

Find all solutions in the interval [0, 2π).

(sin x)(cos x) = 0

Answer 1

Since these are being multiplied, if one of them equals 0, they both do. From the unit circle, the values that make cos 0 are pi/2 and 3pi/2, while for sin, its pi and 2pi

Answer 2

either sinx = 0
or
cosx=0
from 0 to 2pi
sinx = 0 at pi/2 and 3pi/2
cosx =0 at 0 , pi and 2pi
the roots are thus: 0, pi/2, pi, 3pi/2, 2pi

Answer 3

pi divided by two, π

0, pi divided by two border=, π, three pi divided by two

π, three pi divided by two

0, three pi divided by two

(my options)

Answer 4

sorry 2pi isnt included, its open interval

Answer 5

its the second option

Answer 6

Second choice ?

Answer 7

okay cool, thank youu (:

Answer 8

no prob

Answer 9

True

Answer 10

Since these are being multiplied, if one of them equals 0, they both do. From the unit circle, the values that make cos 0 are pi/2 and 3pi/2, while for sin, its pi and 2pi