Posted by alex on .
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,
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,
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,
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