Question

HELP?
FIND ALL SOLUTIONS OF THE EQUATION 4SIN^2X=4COS x+5 IN THE INTERVAL [0,2PI)

Answer 1

\[4sin^2x=4cosx+5\]\[4(1-cos^2x)=4cosx+5\]\[4-4cos^2x=4cosx+5\]\[4cos^2x + 4cosx+1=0\]\[(2cosx+1)^2=0\]\[cosx=-0.5\]Can you solve x?

Answer 2

okay can u please explain step-by-step please? i know its a drag but can u?

Answer 3

It's okay :)
First, you can see that there is a \(sin^2x\) on the left and a cosx on the right. To solve the equation, we'd better change it into one ratio. cosx is difficult to change. So, I changed \(sin^2x\) into \(1-cos^2x\) by using the formula
\(sin^2x + cos^2x =1\)
Got it so far?

Answer 4

YEAH :)

Answer 5

Then, it becomes the quadratic equation... Any questions?

Answer 6

well yeah... why exactly does it become the quad. equation?

Answer 7

like as the next step for this problem

Answer 8

because you can rearrange the terms into the form \(ax^2 + bx + c =0\)
See the fourth line there~

Answer 9

okay now im getting a little thrown off :(

Answer 10

Hmm... which part you don't understand?

Answer 11

ummm welll they part where u say that< the terms could be rearranged into the form ax^2+bx+c=0> why is this equation necessary 4 this problem?

Answer 12

or better yet at this point in the problem?

Answer 13

Because you need to solve it. When you see quadratic question, you can solve it easier..

Answer 14

Actually, it doesn't matter if it is a quadratic equation. As long as you can solve it, that's fine. But in this case, it just comes to be a quadratic equation ...

Answer 15

okay so the quadartic equation comes out 2 b? what exactly.

Answer 16

See the second to the fourth lines :)

Answer 17

okay so line four....which is 4cos^2+4cosx+1=0 this was gathered from subtracting the four and five and then multiplying it by -1 right and moving it over?

Answer 18

\[4-4cos^2x=4cosx+5\]\[4-4cos^2x-4=4cosx+5-4\]\[-4cos^2x=4cosx+1\]\[-4cos^2x+4cos^2x=4cosx+1+4cos^2x\]\[0=4cosx+1+4cos^2x\]\[4cos^2x+4cosx+1=0\]

Answer 19

ok much clearer :] thank u!

Answer 20

welcome :)
Any questions?

Answer 21

yeah after this like there is nothing else? this is the answer for the problem?

Answer 22

Nope :|
you need to solve it!!!
Factor the left side, you'll get \[(2cosx+1)^2=0\]So, \[2cosx+1=0\]Solve x.

Answer 23

so it'll be cosx=-1/2 and then i just find the two references on the unit circle right?

Answer 24

Yes Yes!

Answer 25

k thank u!

Answer 26

welcome :)