1. ## 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

2. 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 $\displaystyle y = \cos (x)$.

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

4. so are these the solutions

y=cos x

(2y-1)(y+1)

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

is this correct

5. 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

6. 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

7. 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

8. 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

9. 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?

10. 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

11. 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)