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March 27, 2017

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Please help I need to solve the following question for x, where 0<x<2pi. and then write a general solution for the equation. The question is 2cos^2 x + cos x =1. I need to solve this using radian measurement. I solved it using algebra and was told I wasn't allowed. Could you please show me the steps to solve it using radian measurement and then how to write a general equation also using radian measurement. Please and thanks

  • trig 30 - ,

    hey Rae,
    there is two ways you can do this...

    1/
    find the common factor of 2cos^2x+cosx=1
    and bring out the front which = cosx(2cosx+1)=1
    therefore either cosx=1 OR 2cosx+1=1 (because there will be multiple answers)

    for cosx=1 use the unit circle to show where cosx=1 which is on the x axis because cos(pi)=1 which is show at points 0 and pi on the unit circle.

    for 2cosx+1=1 simplify to cosx=-1/2
    cos(pi/3)=-1/2 therefore x=pi/3
    graph this in the 2nd and 3rd quadrant (s and t) because cos is negative in those quadrants

    you should now have 4 "lines" on your unit circle graph. start at zero going anti clock wise labelling the points.

    which will be x=0, (pi- pi/3), pi, (pi+pi/3) which simplified will equal
    0, 2pi/3, pi, 4pi/3 these are your four possible answers for x.


    2/ the other way to do this is using the quadratic formula.
    in the equation 2cos^2x+cosx=1 , let cosx=t and substitute that in

    so your equation should now be 2t^2+t=1
    take the 1 over to the other side so it is 2t^2+t-1=0 which gives you a quadratic put this into the quadratic formula

    so it will be now t= 1 +or- the square root (1^2-4*2*-1)all divided by 2*2
    which simplifies to t=(1+or- 3)/4
    therefore t= 1 OR -1/2
    sub cosx=t into that
    therefore cosx=1 OR cosx=-1/2
    its the the same as 1/ from now on..

    hope this helps.:D

  • trig 30 - ,

    I disagree with "belle"

    2cos^2 x + cos x =1
    2cos^2 x + cos x -1 = 0
    (2cosx - 1)(cosx + 1) = 0
    cosx = 1/2 or cosx = -1

    case1: cosx = 1/2
    we know cosπ/6 = 1/2 and the cosine is positive in I and IV
    so x = π/6 or x = 2π - π/6 = 11π/6

    case 2: cosx = -1
    x = π

    so x = π/6 , π, 11π/6

    since the period of cosx is 2π, general solutions would be
    π/6 + 2kπ
    π + 2kπ or π(1+2k)
    11π/6 + 2kπ, where k is an integer.

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