Solving Trigonometric Equations...6

Discussion in 'Geometry and Trigonometry' started by nycmathguy, Dec 18, 2021.

  1. nycmathguy

    nycmathguy

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    Section 5.3

    Screenshot_20211216-022506_Samsung Notes.jpg

    IMG_20211218_181055.jpg
     
    nycmathguy, Dec 18, 2021
    #1
  2. nycmathguy

    MathLover1

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    38.

    sec(x)+tan(x)=1 .........interval [0,2pi]

    1/cos(x)+sin(x)/cos(x)=1

    (1+sin(x))/cos(x)=1

    1+sin(x)-cos(x)=0

    rewrite using identity sin(x)-cos(x)=sqrt(2)sin(x-pi/4)

    1+sqrt(2)sin(x-pi/4)=0

    sqrt(2)sin(x-pi/4)=-1

    sin(x-pi/4)=-1/sqrt(2)

    sin(x-pi/4)=-sqrt(2)/2

    general solutions:

    Radians:
    x-pi/4=5pi/4+2pi*k
    x-pi/4=7pi/4+2pi*k
    solve for x
    x=5pi/4+2pi*k+pi/4 =>x=3pi/2+2pi*k
    x=7pi/4+2pi*k+pi/4=>x=2pi++2pi*k

    solutions for given interval:
    x=3pi/2+2pi*k
    x=2pi++2pi*k
    Degrees:
    x=0°
    x=360°
     
    MathLover1, Dec 19, 2021
    #2
    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    I didn't know the following trigonometric identity:

    sin(x)-cos(x)=sqrt(2)sin(x-pi/4)

    This is why I got stuck. I will do a few more today.
     
    nycmathguy, Dec 19, 2021
    #3
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