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March 26, 2017

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Solve 7 cos(2 theta) = 7 sin^2(theta) + 3 for all solutions 0 less than or equal to theta less than 2pi.






Solve 7 sin(2 w) - 4 cos(w) =0 for all solutions 0 le w lt 2pi

  • precalculus - ,

    le= less than or equal to
    lt = less than

    I keep getting numbers that are not on the unit circle and then when I try to plug them into my calculator I'm getting really big numbers. Please help. I've tried looking in my book but there are no problems like these.

  • precalculus - ,

    7 cos(2 theta) = 7 sin^2(theta) + 3
    7 cos^2 Ø - 7sin^2 Ø = 3
    cos^2Ø - sin^2 Ø = 3/7
    cos 2Ø = 3/7

    so 2Ø is in I or in IV
    set your calculator to radians
    2Ø = 1.1279 or 2Ø = 2π-1.1278 = 5.1553
    Ø = .5639 or Ø = 2.578
    Now the period of cos 2Ø = π , so by adding multiples of π to any answer will yield a new answer

    so Ø = .5639+π = 3.7055
    Ø = 5.1553+π = 5.71924



    7 sin 2w - 4cos w = 0
    7(2sinwcosw) - 4cosw = 0
    cosw(14sinw - 4 ) = 0
    cosw = 0 or sinw = 2/7

    if cosw = 0, then w = π/2 , 3π/2
    if cosw = 2/7 , then w = 1.281 or 2π - 1.281 = 5.002

  • precalculus - ,

    Thank you for your help! My numbers were way off haha.

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