Question

Two x-interecepts of the function h(x) = -2sin(x- π/2) are

Answer 1

solve for x
-2sin(x- π/2) = 0

Answer 2

with the sin in there do i do sin-1 instead of regular sin to figure it out?

Answer 3

the curve have been shifted right by a distance of pi/2

the normal root for -2sin(x) is the origin, then pi, 2pi etc

the phase shift right will see the intercepts move to

-pi/2, pi2, 3pi/2, 5pi/2 etc

Answer 4

or,
\[ \sin ( x - \pi/2) = 0 \]
or, \[ \sin(x- \pi/2) = \sin 0 = \sin \pi = \sin 2\pi = \sin 3pi = \sin 4\pi = ... \]
\[ x-\pi/2 = 0 = \pi = 2\pi = ... \]

Answer 5

π/3 and 5π/2 would that be the answer

Answer 6

find any two values of x

Answer 7

nvm answer would be 0 and 2pi?

Answer 8

plugin those values... into woflramalpha, and see if you get 0 or not