Posted by alex on Monday, April 7, 2008 at 9:02am.
the problem is
2cos^2x + sinx-1=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:
cos2x + sinx =0
1 - 2sin^2x + sinx = 0
I think i'm supposed to factor it next but i'm not sure?
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
replace cos^2x by 1-sin^2x
2(1-sin^2x) + sinx -1 = 0
2-2sin^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 = 30º)
x must be in quadrant III or IV
x = 180+30 =210 or
x = 360-30 = 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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