Question

Find all solutions to the equation.

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

Answer 1

https://www.wolframalpha.com/input/?i=7+sin%5E2%28x%29+-+14+sin+x+%2B+2+%3D+-5

I don't know if that will be of any help, but there you go

Answer 2

this is a quadratic in terms of sin(x)

Answer 3

Do you know how to solve the following for u:
\[7u^2-14u+2=-5\]

Answer 4

bring five over and equal it to zero?

Answer 5

7u^2-14u-3??

Answer 6

but 2+5 is 7 not -3

Answer 7

\[7u^2-14u+2=-5 \\ \text{ add 5 on both sides } \\ 7u^2-14u+2+5=-5+5 \\ 7u^2-14u+7=0\]

Answer 8

oo ok so now do we combine 7u^2 with 14u?

Answer 9

you can't
they aren't like terms

Answer 10

have you ever solved a quadratic before

Answer 11

you can try factoring or use the quadratic formula....or completing the square

Answer 12

ok lets do quadratic form is A=7u^2 and b=14U and c=7 ?

Answer 13

do you mean a=7, b=-14,c=7 ?

Answer 14

i got 7=1 for posititive quardatic

Answer 15

I have no clue how you can conclude 7 is 1...

Answer 16

7 is greater than 1
so it cannot be 1

Answer 17

\[au^2+bu+c=0 \\ u=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 18

all you do is enter in the a,b,c into the above formula

Answer 19

yes i know the rhyme for the quad. o got 1 for positive

Answer 20

i got 1=1

Answer 21

what did you get for u ?

Answer 22

i got u equals 1 for both eqautions

Answer 23

\[u=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \text{ we have } a=7,b=-14,c=7 \\ u=\frac{-(-14) \pm \sqrt{(-14)^2-4(7)(7)}}{2(7)} \\ u=\frac{14 \pm \sqrt{196-196}}{14} \\ u=1\]oh so you mean u=1
not 1=1..which doesn't really help us

Answer 24

ok well remember u was in the place of sin(x)

Answer 25

sin(x)=1 when x=?

Answer 26

1?

Answer 27

sin(1) is not 1

Answer 28

look at the unit circle if you have to
for what angles on the unit circle do you have the y-coordinate is 1 on the unit circle

Answer 29

90 deg

Answer 30

yep so
90 deg+360 deg*k
where k is a integer

Answer 31

or you prefer....
pi/2+2pi*k
where k is an integer

Answer 32

is that the answer ?

Answer 33

yep one of those are the answer
it depends on the preference of your teacher

Answer 34

wow thats was alot easier than i expected. when can you not use the quadratic form?

Answer 35

when it can't be written in the form au^2+bu+c=0

Answer 36

if it can be written in the form au^2+bu+c=0
then it is in quadratic form in terms of u
whether u be sin(x) or cos(x) or tan(x)...

Answer 37

ok thank you so much