Posted by Joanie on Tuesday, January 20, 2009 at 6:13pm.
Solve 4sin^2x + 4(squareroot of 2)cosx -6 for all real values of x.
- Math - Reiny, Tuesday, January 20, 2009 at 6:47pm
Did you mean
4sin^2x + 4√2cosx -6 = 0 ?
if so then
4(1-cos^2x) + 4√2cosx - 6 = 0
4 - cos^2x + 4√2cosx - 6 = 0
2cos^2x + 2√2cosx + 1 = 0
let cosx = y
then y = (-2√2 ± √(8 - 4(2)(1))/4
y = -√2/2
then cosx = -√2/2
we know that cos 45 = √2/2
and x must be in the II or III quadrant, so
x = 180-45 = 135 degrees or
x = 180+45 = 225 degrees
in radians that would be 3pi/4 or 5pi/4 radians
x = 135 + 360k, 225 + 360k, k an integer
x = 3pi/4 + 2kpi, 5pi/4 + 2kpi
- Math - Joanie, Tuesday, January 20, 2009 at 8:34pm
Thank you for helping me with this problem.
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