Posted by alex on Monday, April 7, 2008 at 9:02am.
the problem is
2cos^2x + sinx1=0
the directions are to "use an identity to solve each equation on the interval [0,2pi).
This is what i've done so far:
2cos^2x+sinx1=0
2cos^2x1+sinx=0
cos2x + sinx =0
1  2sin^2x + sinx = 0
2sin^2x+sinx1=0
2sin^2xsinx+1=0
I think i'm supposed to factor it next but i'm not sure?
(2sinx?)(sinx?)
would doing that step be correct?

math  alex, Monday, April 7, 2008 at 9:05am
although, after looking at this problem i'm still not sure why i had to change the original problem? I've just been following the notes that were given to me by my teacher and that's what she has done but why do i have to change the original problem it seems like just a bunch of extra work

math  Dena, Monday, April 7, 2008 at 9:28am
2cos^2x+sinx1=0
replace cos^2x by 1sin^2x
2(1sin^2x) + sinx 1 = 0
22sin^2x +sinx 1 = 0
2sin^2x+sinx+1=0
2sin^2x  sinx  1=0
(2sinx + 1)(sinx 1)=0
we will get
sinx = 1/2 and 1

math  Reiny, Monday, April 7, 2008 at 9:44am
carrying on where Dena left off....
sinx = 1/2 (angle in standard postition = 30ยบ)
x must be in quadrant III or IV
x = 180+30 =210 or
x = 36030 = 330
if sinx = 1
x = 90
so in degrees x = 90, 210 or 330
or in radians
x = pi/2, 7pi/6 or 11pi/6
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