Question

!!!!!can someone explain to me how to do this kind of precalculus problem pleaseee****

Answer 1

You pass the cos to the left side, so arcsin(-1)

Answer 2

Then solve it as an easy algebra question

Answer 3

arcsin? we never learned that kind of way have time to explain?

Answer 4

can you answer this: cos (???) = -1

what replaces ???

Answer 5

arcsin is an alternative saying for sin^-1

Answer 6

cos(pi)

Answer 7

And for period for a cos function, it's 2pi/|b|

Answer 8

but its not sin^-1 or cos^-1..?

Answer 9

Oh yeah, arccos >.>

Answer 10

(x - pi/9) = arccos(-1)
(x - pi/9) = pi

Isolate x

x = pi + pi/9= 10pi/9

As for the period, we know that sin and cos function's period could be found using 2pi/|b|
b is one, 2pi/|1| = 2pi

So the answer would be x = 10pi/9 + 2pin

Answer 11

how do we know that (x-pi/9) = arccos(-1) which in turn equals pi? is that for any cos (x-a pi something?)

Answer 12

Have you tried to watch the videos that come with your online course?

Answer 13

We know that 90 degrees is pi radian
cos(90) = -1
therefore cos(pi) = -1

If arccos(-1) = 90 degrees and 90 degrees = pi radian
then arccos(pi)