Question

2cos^2x+sinx-2=0
Solve the equation on the interval [0,2pi]

Answer 1

@Directrix

Answer 2

Write cos^2x in (1-sin^2x)

Answer 3

What next ?

Answer 4

Let S = sinx

Answer 5

What does that mean?

Answer 6

Well, firstly we have \(2(1-\sin^2x)+\sin x-2=0\)

Answer 7

Now we have \(2(1-S^2)+S-2=0\)

Answer 8

Do you understand?

Answer 9

Yes

Answer 10

Then solve the quadratic equation

Answer 11

Disturbute 2?

Answer 12

yep

Answer 13

I stopped at \[2\sin(x)^2+\sin(x)\]

Answer 14

Well I thought it's an equation

Answer 15

Meaning?

Answer 16

where's the equal sign

Answer 17

2sin(x)^2+sin(x)=0

Answer 18

it should be - instead of +

Answer 19

Why?

Answer 20

Oh ok

Answer 21

then solve that equation

Answer 22

You have \(S^2-2S=0\) right?

Answer 23

yes

Answer 24

then solve this quadratic equation

Answer 25

Factor?

Answer 26

yep

Answer 27

\[sinx(1-2sinx)=0\]

Answer 28

Yep

Answer 29

Sinx=0

Answer 30

Sinx=-1/2

Answer 31

yep

Answer 32

well, not -1/2 but +1/2

Answer 33

It'll be negative

Answer 34

why?

Answer 35

1-2sinx=o

Answer 36

That's how

Answer 37

well, 1-2sinx=0, 1=2sinx, 1/2=sinx

Answer 38

When you the put the one to the right it becomes negative

Answer 39

1-2sinx=0
1-2sinx+2sinx = 0+2sinx
1=2sinx

Answer 40

What would be the x

Answer 41

well, sinx=0 or sinx=1/2

Answer 42

what can x be?

Answer 43

Pi, pi/6

Answer 44

there should be 5 answers

Answer 45

they're 0,pi,2pi and pi/6,5pi/6

Answer 46

i got to go now