Trigonometry

posted by .

Find all solutions of 4(cos(x))**2-4 sin(x)-5=0 in the interval (6pi, 8pi)

I tried to work it out and got: 4(cos**2x)-4cosx -9 = 0, but I can't figure out what cosx = from there to finish the problem.

  • Trigonometry -

    ________________________________________

    Remark :

    ( cos x ) ^ 2 = 1 - ( sin x ) ^ 2
    ________________________________________


    4 ( cos x ) ^ 2 - 4 sin( x ) - 5 = 0

    4 [ 1 - ( sin x ) ^ 2 ] - 4 sin( x ) - 5 = 0

    4 - 4 ( sin x ) ^ 2 - 4 sin ( x ) - 5 = 0

    - 4 ( sin x ) ^ 2 - 4 sin ( x ) - 1 = 0 Multiply both sides by - 1

    4 ( sin x ) ^ 2 + 4 sin ( x ) + 1 = 0
    ________________________________________

    Remark :

    ( a + b ) ^ 2 = a ^ 2 + 2 a b + b ^ 2

    So :

    4 sin ( x ) ^ 2 + 4 sin ( x ) + 1 = [ 2 sin ( x ) + 1 ] ^ 2

    Becouse :

    [ 2 sin ( x ) + 1 ] ^ 2 =

    [ 2 sin ( x ) ] ^ 2 + 2 * 2 sin ( x ) + 1 ^ 2 =

    4 sin ( x ) ^ 2 + 4 sin ( x ) + 1


    4 ( sin x ) ^ 2 + 4 sin ( x ) + 1 = 0 is same :

    [ 2 sin ( x ) + 1 ] ^ 2 = 0
    ________________________________________

    [ 2 sin ( x ) + 1 ] ^ 2 = 0 Take the square root of both sides

    2 sin ( x ) + 1 = 0 Subtract 1 from both sides

    2 sin ( x ) = - 1 Divide bboth sides by 2

    sin ( x ) = - 1 / 2


    sin ( x ) = - 1 / 2 for :

    x = 7 pi / 6 = 210 °

    and

    x = 11 pi / 6 = 330 °


    The period of sin ( x ) is 2 pi

    So :

    sin ( x ) = - 1 / 2 for :

    x = 2 n pi +7 pi / 6

    and

    x = 2 n pi + 11 pi / 6

    n is an integer


    In the interval ( 6 pi, 8 pi )

    x = 2 * 3 pi + 7 pi / 6

    and

    x = 2 * 3 pi + 11 pi / 6

    Solutions :

    x = 6 pi + 7 pi / 6

    and

    x = 6 pi + 11 pi / 6


    OR

    x = 6 * 6 pi / 6 + 7 pi / 6 = 36 pi / 6 + 7 pi / 6 = 43 pi / 6

    and

    x = 6 * 6 pi / 6 + 11 pi / 6 = 36 pi / 6 + 11 pi / 6 = 47 pi / 6

  • Trigonometry -

    You apparently think that sinx = 1+cosx
    The correct identity is cos^2 + sin^2 = 1
    so, cos^2 x = 1-sin^2 x

    4(1-sin^2 x) - 4sinx - 5 = 0
    4 - 4sin^2 x - 4sinx - 5 = 0

    4sin^2 x + 4sinx + 1 = 0
    (2sinx + 1)^2 = 0
    2sinx + 1 = 0
    sinx = -1/2

    Take it from there.

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-Calculus

    How can this identity be proved/verified?
  3. Precal

    I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1 - sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 = - …
  4. maths - trigonometry

    I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the second. …
  5. pre calc trig check my work please

    sin x + cos x -------------- = ? sin x sin x cos x ----- + ----- = sin x sin x cos x/sin x = cot x this is what i got, the problem is we have a match the expression to the equation work sheet and this is not one of the answers. need
  6. trig

    Find all solutions of the equation. Leave answers in trigonometric form. x^5-1024=0 I got 4(cos tehta + i sin tehta), tehta = 0, 2pi/5, 4pi/5, 6pi/5, 8pi/5 is this right
  7. Trigonometry

    Find all solutions of 4 (cos (x)**2)-1=0 in the interval (6pi, 8pi). (Leave your answers in exact form and enter them as a comma-separated list.)
  8. Math, please help

    Which of the following are trigonometric identities?
  9. Math - trig

    Determine the general solution of 8 cos^2x - 4cosx - 1 =0 (8cos^2x - 4cosx)/8 =1/8 cos^2x - 4cosx=1/8 cos^2x/cosx =1+4cosx/cosx cosx = 5 x = cos^-1(5) x = error
  10. Trigonometry

    Solve the equation for solutions in the interval 0<=theta<2pi Problem 1. 3cot^2-4csc=1 My attempt: 3(cos^2/sin^2)-4/sin=1 3(cos^2/sin^2) - 4sin/sin^2 = 1 3cos^2 -4sin =sin^2 3cos^2-(1-cos^2) =4sin 4cos^2 -1 =4sin Cos^2 - sin=1/4 …

More Similar Questions