Pre-calculus

posted by .

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 -

    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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. pre-calc

    Solve: cos(2x-180) - sin(x-90)=0 my work: cos2xcos180 + sin2xsin180= sinxcos90 - sin90cosx -cos2x - sin2x= cosx -cos^2x + sin^2x -2sinxcosx=cosx I'm stuck here. I tried subtracting cosx from both sides and making sin^2x into 1- cos^2x, …
  2. pre-cal

    Simplify the given expression........? (2sin2x)(cos6x) sin 2x and cos 6x can be expressed as a series of terms that involve sin x or cos x only, but the end result is not a simplification. sin 2x = 2 sinx cosx cos 6x = 32 cos^6 x -48
  3. Calculus

    Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 …
  4. Math/Calculus

    Solve the initial-value problem. Am I using the wrong value for beta here, 2sqrt(2) or am I making a mistake somewhere else?
  5. Calculus

    Find the derivative of cos(sqrt(e^(x^3)cos(x)) I got -sin(sqrt(e^(x^3)cos(x))*((sqrt(e^(x^3)cos(x)) Do I just leave the e^(x^3)cosx alone since it's "e" or do I still have the find the derivative for it?
  6. Calc II

    For the following question, we need to find the length of the polar curve: r= 2/(1-cosx) from pi/2 </= x </= pi dr/dx= -2sinx/(1-cosx)^2 therefore the length would be expressed as follows: the integral from pi/2 to pi of (sqrt)[(2/(1-cos))^2 …
  7. trigonometry (please double check this)

    Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only. 1. sin2ƒÆ = (sqrt 3)/2 2. sin^2ƒÆ = …
  8. Calc.

    Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= …
  9. calculus

    Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= …
  10. Calculus

    Integrate 1/sinx dx using the identity sinx=2(sin(x/2)cos(x/2)). I rewrote the integral to 1/2 ∫ 1/(sin(x/2)cos(x/2))dx, but I don't know how to continue. Thanks for the help. Calculus - Steve, Tuesday, January 12, 2016 at 12:45am …

More Similar Questions