Wednesday

January 28, 2015

January 28, 2015

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

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 12:03pmif 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

**Answer this Question**

**Related Questions**

Calculus - Please look at my work below: Solve the initial-value problem. y'' + ...

calculus - Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x...

Math/Calculus - Solve the initial-value problem. Am I using the wrong value for ...

Calc. - Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln...

math - Prove that for all real values of a, b, t (theta): (a * cos t + b * sin t...

Math - Calculus - The identity below is significant because it relates 3 ...

Precal - I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A...

math - A trigonmetric polynomial of order n is t(x) = c0 + c1 * cos x + c2 * cos...

Calculus - Second Order Differential Equations - Posted by COFFEE on Monday, ...

Precalculs - I have no idea how to do these type of problems. -------Problem...