Monday

July 28, 2014

July 28, 2014

Posted by **alex** on Monday, April 7, 2008 at 9:02am.

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:

2cos^2x+sinx-1=0

2cos^2x-1+sinx=0

cos2x + sinx =0

1 - 2sin^2x + sinx = 0

-2sin^2x+sinx-1=0

2sin^2x-sinx+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 9:05amalthough, 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 9:28am2cos^2x+sinx-1=0

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

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 9:44amcarrying 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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