1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solve acos²θ+bsinθ+c=0 for all values 0≤θ≤360°

  1. Dec 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Solve acos²θ+bsinθ+c=0 for all values 0≤θ≤360°

    a=16
    b=6
    c=-12

    So 16cos²θ+6sinθ-12=0

    2. Relevant equations
    Cos²x=1-Sin²x

    3. The attempt at a solution
    Identity: Cos²x=1-Sin²x

    16(1-Sin²θ)+6Sinθ-12=0

    16-16Sin²θ+6Sinθ-12=0

    6Sinθ-16Sin²θ=12-16=-4

    Divide by 2(?)

    3Sinθ-8Sin²θ=-2
    3Sinθ-4Sin³θ=-2

    3Sinθ-4Sin³θ=Sin3θ

    Sin3θ=-2

    I realised that I am stuck at this point and can’t work out how to find θ from this, although it feels like I am missing something obvious. Any help would be appreciated!
     
  2. jcsd
  3. Dec 6, 2016 #2
    $$
    3\sinθ-8\sin^2θ=-2$$
    $$3\sinθ-4\sin^{\color{red}{3}}θ=-2

    $$

    Where does that three come from ?
     
  4. Dec 6, 2016 #3
    I tried to change 8Sin²θ into Sin³θ, but I guess that doesn’t work. I think it’s because I had Sin3θ written down, so I was trying to make it work out as that.
     
  5. Dec 6, 2016 #4
    16-16Sin²θ+6Sinθ-12=0

    16Sin²θ-6Sinθ=4

    Sinθ(16Sinθ-6)=4

    I still can’t work out what to do next.
     
  6. Dec 6, 2016 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Hint: Let sinθ = x. What kind of equation would you have for x?
     
  7. Dec 6, 2016 #6
    A quadratic equation?

    ax²+bx+c=0

    16x²-6x-4=0

    Sinθ=6±√-6²-(4×16×-4)∕2×16 (Sorry not sure how to type that so it looks more clear)

    Sinθ=0.65
    Sinθ=-0.28

    Sin–1(0.65)=40.54°
    Sin–1(-0.28)=-16.26°

    But seeing as it is between 0° and 360°

    θ1 = 40.54°
    θ2 = 180°-40.54°=139.46° (?)
     
  8. Dec 6, 2016 #7
    Hopefully that's right
     
  9. Dec 6, 2016 #8
  10. Dec 6, 2016 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You wrote
    $$\sin \theta = 6 \pm \frac{\sqrt{-6^2 - (4 \times 16 \times -4)}}{2} \times 16 $$
    Did you mean
    $$ \sin \theta = \frac{6 \pm \sqrt{(-6)^2 - 4\times16\times (-4)}}{2 \times 16}?\:$$
    If so, use parentheses, like this:
    [6 ± √[(-6)^2 - 4 × 16 ×(-4)]]/(2 × 16).
    Better still, use LaTeX.

    Anyway, ##x = 0.65## and ##x = -0.28## are nowhere near being solutions of the equation ##16 x^2 - 6 x -4 = 0##. In fact, if ##f(x) = 16 x^2 - 6 x - 4## we have ##f(0.65) = -1.1400## and ##f(-0.28) = -1.0656##.
     
  11. Dec 6, 2016 #10
    Yes I meant that, sorry this is the first time I have tried to type an equation rather than write it.

    Not sure what you mean there. It asks for the angles between 0° and 360°, can I not just Sin^-1(x) to get θ?
     
  12. Dec 6, 2016 #11

    Doc Al

    User Avatar

    Staff: Mentor

    First you must correctly solve the quadratic. Then you can solve for all values of θ that work.
     
  13. Dec 6, 2016 #12

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You obtained ##x=\sin \theta## by solving a quadratic equation; then you used the values of ##x## to get values of ##\theta##. Your values of ##x## are incorrect, so your values of ##\theta## are also incorrect.

    Do not take my word for it: test it for yourself. Take your value ##\theta = 40.54^o## and plug it into the function ##f(\theta) = 16 \cos^2(\theta) + 6 \sin(\theta) -12##, to see if you get zero (except for small roundoff errors).
     
    Last edited: Dec 6, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Solve acos²θ+bsinθ+c=0 for all values 0≤θ≤360°
  1. Solve for tan(θ) (Replies: 9)

Loading...