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

general solution

x = 135 + 360k, 225 + 360k, k an integer

or

x = 3pi/4 + 2kpi, 5pi/4 + 2kpi

- Math -
**Joanie**, Tuesday, January 20, 2009 at 8:34pm
Thank you for helping me with this problem.

Joanie

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