Solutions with intervals

• Jun 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
• Jun 30th 2008, 03:18 PM
Plato
Make these changes.
$\displaystyle \begin{array}{l} 2\sin ^2 (x) = 2 + \cos (x) \\ 2 - 2\cos ^2 (x) = 2 + \cos (x) \\ 2\cos ^2 (x) + \cos (x) = 0 \\ \end{array}$

Can you solve $\displaystyle 2y^2 + y = 0$?
Let $\displaystyle y = \cos (x)$.
• Jun 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)
• Jun 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
• Jun 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) $\displaystyle \sin^2 x + \cos^2 x = 1$. that means you can change $\displaystyle \sin^2 x$ to $\displaystyle 1 - \cos^2 x$, which is what Plato did
• Jun 30th 2008, 08:00 PM
NoAsherelol
Quote:

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

is my factoring correct
• Jun 30th 2008, 08:11 PM
Jhevon
Quote:

Originally Posted by NoAsherelol
is my factoring correct

no

$\displaystyle 2y^2 + y = 0$

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

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
• Jun 30th 2008, 08:18 PM
NoAsherelol
Quote:

Originally Posted by Jhevon
no

$\displaystyle 2y^2 + y = 0$

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

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
• Jun 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?
• Jul 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
• Jul 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 $\displaystyle \cos x = 0 \implies x = \frac {\pi}2 + k \pi$ for $\displaystyle k \in \mathbb{Z}$

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

(cosine is negative in the second and third quadrants)