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Math Help - please help solve for θ

  1. #1
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    please help solve for θ

    Solve for θ if 2 tan^2 θ + secθ − 4 = 0 0 ≤ θ ≤ 360

    tan^2 θ =sec^2 θ - 1

    therefore 2(sec^2θ -1)+sec θ -4=0

    2sec^2 θ + sec θ -6 =0
    (sec θ + 2) (2sec θ -3)=0

    sec θ = 2 or 2secθ =3

    this is the furthest i got
    please help
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  2. #2
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    Quote Originally Posted by joey1 View Post
    Solve for θ if 2 tan^2 θ + secθ − 4 = 0 0 ≤ θ ≤ 360

    tan^2 θ =sec^2 θ - 1

    therefore 2(sec^2θ -1)+sec θ -4=0

    2sec^2 θ + sec θ -6 =0
    (sec θ + 2) (2sec θ -3)=0

    sec θ = 2 or 2secθ =3

    this is the furthest i got
    please help
    2\tan^2{\theta} + \sec{\theta} - 4 = 0.

    Now, remembering that \tan^2{\theta} + 1 = \sec^2{\theta}, this gives...

    2(\sec^2{\theta} - 1) + \sec{\theta} - 4 = 0

    2\sec^2{\theta} - 2 + \sec{\theta} - 4 = 0

    2\sec^2{\theta} + \sec{\theta} - 6 = 0

    2\sec^2{\theta} + 4\sec{\theta} - 3\sec{\theta} - 6 = 0

    2\sec{\theta}(\sec{\theta} + 2) - 3(\sec{\theta} + 2) = 0

    (\sec{\theta} + 2)(2\sec{\theta} - 3) = 0

    Case 1:

    \sec{\theta} + 2 = 0

    \sec{\theta} = -2

    \cos{\theta} = -\frac{1}{2}

    \theta = \left\{120^\circ, 240^\circ\right\}.

    Case 2:

    2\sec{\theta} - 3 = 0

    2\sec{\theta} = 3

    \sec{\theta} = \frac{3}{2}

    \cos{\theta} = \frac{2}{3}

    \theta = \left\{\arccos{\frac{2}{3}}, 360^\circ - \arccos{\frac{2}{3}}\right\}.
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  3. #3
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    Quote Originally Posted by joey1 View Post
    Solve for θ if 2 tan^2 θ + secθ − 4 = 0 0 ≤ θ ≤ 360

    tan^2 θ =sec^2 θ - 1

    therefore 2(sec^2θ -1)+sec θ -4=0

    2sec^2 θ + sec θ -6 =0
    (sec θ + 2) (2sec θ -3)=0

    sec θ = - 2 or 2secθ =3

    this is the furthest i got
    please help
    hi good job so far ,

    i will use b for convenience

    cos b=2/3 , cos b=-1/2 since cos is reciprocal of sec .

    cos b=2/3 , cos is positive here so it will be in the 1st and 4th quadrant .

    b=48.19 or 311.81

    cos b=-1/2 , cos -ve so 2nd and 3rd quadrant .

    and we know from the special angles cos 60 =1/2

    so b=120 or 240

    therefore b=48.19 , 120 , 240 , 311.81
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  4. #4
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    Quote Originally Posted by joey1 View Post
    Solve for θ if 2 tan^2 θ + secθ − 4 = 0 0 ≤ θ ≤ 360

    tan^2 θ =sec^2 θ - 1

    therefore 2(sec^2θ -1)+sec θ -4=0

    2sec^2 θ + sec θ -6 =0
    (sec θ + 2) (2sec θ -3)=0

    sec θ = 2 or 2secθ =3

    this is the furthest i got
    please help
    Just continue form there as IF (x + 2)(2x - 3) = 0, THEN x = -2 or 3/2.

    So if sec θ = 2, you can find θ directly, and likewise if sec θ = 3/2.

    A small word of advice: There is no need to convert to primary functions as you should learn to as readily use the so called secondary functions. I prefer to look at all six as exactly equivalent in usefulness as they stand. The division into primary and secondary is useful for those having difficulty memorising, but it pays off to look as all six ratios equally well. This is particularly so when applying to "life" problems and there is no time spent on translating back and forth.
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