Question

I have no clue here please help..
f(x)=8−6sinπ5(x+1)Find the first negative value of x for which f(x) = 3

Answer 1

let f(x) = 3 and algebra thru it

Answer 2

f(x)=8−6sinπ5(x+1)

3=8−6sinπ5(x+1)

-5=−6sinπ5(x+1)

5/6= sinπ5(x+1)

i got no idea what the rest of that is spose to mean

Answer 3

\[ maybe:~\sin~[5π(x+1)]\]
if so then inverse

arcsin(5/6) = 5π(x+1)

arcsin(5/6)/5π = x+1

arcsin(5/6)/5π - 1 = x

Answer 4

do you have any idea what the answer would be

Answer 5

I've been working on this for close to an hour and im no where near getting it

Answer 6

apart from inputing that into a calculator ... no

Answer 7

using a derivative, i know that at -.9 there is the first miminum point

using the calculations above i get about -.93729, which is to far

the difference between them is .03729

-.90000
.03729
--------
- .86271 gets rather close