Solve 4sin^2x + 4(squareroot of 2)cosx -6 for all real values of x.

Answer 1

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

Answer 2

Thank you for helping me with this problem.

Joanie