Question

Find all solutions to the equation.

7 sin^2x - 14 sin x + 2 = -5

Answer 1

can you solve
7a^2-14a+2=-5?

Answer 2

isn't it 7a^2 -14a +7 = 0 now?

Answer 3

Hi Death :) thanks for viewing my question

Answer 4

well im here to help now. sorry for leaving earlier. had to finish some summer school homework

Answer 5

No problem! Thank you so much!

Answer 6

I hate my Pre-Calc course...the text book is the freaking online textbook is the most difficult thing to navigate ever...plus it's a college course while I'm in high school math! I have no choice in the matter...damn pre-calc...

Answer 7

its either ''a = -0.286 - 0.958'' or ''a = -0.286 + 0.958".

Answer 8

I feel you. I hate online class textbooks

Answer 9

so by either do you mean both or you're not sure? I don't THINK I have to use unit circle for this one...probably...

Answer 10

because it doesn't give me an interval

Answer 11

wait? no interval?

Answer 12

7a^2 -14a +7 = 0
yes. solve this
7(a-1)^2 = 0
so a = 1
so sin(x) =1
this is at pi/2,6pi/2,10pi/2 = pi/2+2pi*k where k is an integer

Answer 13

there are infinite solutions, no interval needed.

Answer 14

no, no interval...just all possible solutions

Answer 15

7a^2 -14a +7 = 0
yes. solve this
7(a-1)^2 = 0
so a = 1
so sin(x) =1
this is at pi/2,5pi/2,9pi/2
so solutions are pi/2+2pi*k where k is an integer

Answer 16

where did k come from? Ummm I got confused...

Answer 17

he means the "k" should be the integer

Answer 18

well sin(x) = 1 at pi/2 and also (since we are not restricted to an interval) at 5pi/2
and 9pi/2
so
{....,pi/2,5pi/2,9pi/2,......}
another way to write this is
pi/2+2pi*k where k is an integer

Answer 19

for k=0
you get pi/2+2pi*0 = pi/2
for k = 1
you get pi/2+2pi*1 = 5pi/2
for k = 2
you get 9pi/2
.......

Answer 20

so if you let k run through all the integers {...-3,-2,-1,0,1,2,3.....} you will get "all" of the soltions

Answer 21

I know this is allot, but I think you should spend some time and make sure you understand everything I said. It will help you allot in this class..

Answer 22

okay thank you very much for all your time