Posted by **Lucy** on Monday, October 13, 2008 at 11:42am.

solve 4 sin^2x + 4 sqrt 2 cos x-6=0 for all real values of x.

Going by the example in the book I got to: 4 cos^2x + 4sqrt2 cosx -10 = 0 but do not know how to proceed.

Any help would be great.

Thanks

- Pre-calculus -
**Reiny**, Monday, October 13, 2008 at 12:03pm
if you replaced sin^2x with 1 - cos^2x and simplified correctly you should have had

4cos^2x - 4√2cosx + 2 = 0

solve this as a quadratic using the formula to get

cosx = (4√2 ± 0)/8 = √2/2

at this point you should realize that the equation would have factored to

(2cosx - √2)^2 = 0

then 2cosx = √2

and cosx = √2/2

so x = 45º or 315º (or pi/2 and 7pi/2 radians)

since it asked for all real values of x, we could give a general solution of

45º + 360kº or 135º + 360kº where k is an integer.

I will leave it up to you to give the general solution in radians

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