Solutions with intervals

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June 30th 2008, 03:03 PM
NoAsherelol

Solutions with intervals

Find all solutions in the interval [0, 2pi)

2sin^2 x = 2 + cox x

I kno that most of the solutions are gonna be 1st quadrant solutions,

sin is positive in the 1st and 2nd and cos is negative in the 4th quadrant
June 30th 2008, 03:18 PM
Plato

Make these changes.
$\begin{array}{l} 2\sin ^2 (x) = 2 + \cos (x) \\ 2 - 2\cos ^2 (x) = 2 + \cos (x) \\ 2\cos ^2 (x) + \cos (x) = 0 \\ \end{array}$

Can you solve $2y^2 + y = 0$ ?
Let $y = \cos (x)$ .
June 30th 2008, 03:26 PM
NoAsherelol

yes i could solve that thank you fo rthe asistance, i dont get where your getting
2-2cos^2 x + 2 + cos(x)
June 30th 2008, 03:45 PM
NoAsherelol

so are these the solutions

y=cos x

(2y-1)(y+1)

Cos(x)=1/2 and cos(x)=-1

is this correct
June 30th 2008, 03:47 PM
Jhevon

Quote:

Originally Posted by NoAsherelol

yes i could solve that thank you fo rthe asistance, i dont get where your getting
2-2cos^2 x + 2 + cos(x)

he used the identity (that you should be familiar wit) $\sin^2 x + \cos^2 x = 1$ . that means you can change $\sin^2 x$ to $1 - \cos^2 x$ , which is what Plato did
June 30th 2008, 08:00 PM
NoAsherelol

Quote:

Originally Posted by Jhevon

he used the identity (that you should be familiar wit) $\sin^2 x + \cos^2 x = 1$ . that means you can change $\sin^2 x$ to $1 - \cos^2 x$ , which is what Plato did

is my factoring correct
June 30th 2008, 08:11 PM
Jhevon

Quote:

Originally Posted by NoAsherelol

is my factoring correct

no

$2y^2 + y = 0$

$\Rightarrow y(2y + 1) = 0$

by the way, you could check whether or not you factored right by multiplying out what you got. you would notice you end up with three terms, as opposed to two, which is what you started with
June 30th 2008, 08:18 PM
NoAsherelol

Quote:

Originally Posted by Jhevon

no

$2y^2 + y = 0$

$\Rightarrow y(2y + 1) = 0$

by the way, you could check whether or not you factored right by multiplying out what you got. you would notice you end up with three terms, as opposed to two, which is what you started with

Im coming up with cos(x)=0 and cos(x)=-1, but the interval is [0, 2pi) with those answers it could only be 1 solution but it asked me to find solutions yo i suck at this class, i might have tell my guidance counselor to switch me out.....sorry if im buiggin the mess out of you
June 30th 2008, 08:29 PM
Jhevon

Quote:

Originally Posted by NoAsherelol

Im coming up with cos(x)=0 and cos(x)=-1, but the interval is [0, 2pi) with those answers it could only be 1 solution but it asked me to find solutions yo i suck at this class, i might have tell my guidance counselor to switch me out.....sorry if im buiggin the mess out of you

the solutions are cos(x) = 0 and cos(x) = -1/2... now, how would we find x?
July 1st 2008, 03:14 AM
NoAsherelol

Quote:

Originally Posted by Jhevon

the solutions are cos(x) = 0 and cos(x) = -1/2... now, how would we find x?

well i know cos is negative in the 2nd and fourth quadrants,

so possible solutions are x=2pi/3 and x=11pi/6
July 1st 2008, 02:23 PM
Jhevon

Quote:

Originally Posted by NoAsherelol

well i know cos is negative in the 2nd and fourth quadrants,

so possible solutions are x=2pi/3 and x=11pi/6

from the graph of cos(x) we know that $\cos x = 0 \implies x = \frac {\pi}2 + k \pi$ for $k \in \mathbb{Z}$

from the table i gave you in one of your other posts and your knowledge of reference angles, we know $\cos x = - \frac 12 \implies x = \frac {2 \pi}3 + k \pi \mbox{ or } \frac {4 \pi}3 + n \pi$ for $n,k \in \mathbb{Z}$

(cosine is negative in the second and third quadrants)