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

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- Jun 30th 2008, 03:03 PMNoAsherelolSolutions 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 PMPlato
Make these changes.

Can you solve ?

Let . - Jun 30th 2008, 03:26 PMNoAsherelol
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 PMNoAsherelol
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 PMJhevon
- Jun 30th 2008, 08:00 PMNoAsherelol
- Jun 30th 2008, 08:11 PMJhevon
- Jun 30th 2008, 08:18 PMNoAsherelol

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 PMJhevon
- Jul 1st 2008, 03:14 AMNoAsherelol
- Jul 1st 2008, 02:23 PMJhevon