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anonymous

  • one year ago

solve for 2cos^2theta + 13costheta = -6. theta is >= 0 and < 360 degrees

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  1. anonymous
    • one year ago
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    \[2\cos^2(\theta) + 13\cos(\theta) = -6\] Like that?

  2. barrycarter
    • one year ago
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    If like ab854 asked, hint: quadratic equation.

  3. anonymous
    • one year ago
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    yes like that, how did you do the theta sign?

  4. anonymous
    • one year ago
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    thank you guys.

  5. anonymous
    • one year ago
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    \[2 \cos ^2\theta+13\cos \theta+6=0\] \[2 \cos ^2\theta+12\cos \theta+\cos \theta+6=0\] \[2\cos \theta \left( \cos \theta+6 \right)+1\left( \cos \theta+6 \right)=0\] \[\left( \cos \theta+6 \right)\left( 2\cos \theta+1 \right)=0\] \[Either~\cos \theta=-6 ~(rejected),~as \left| \cos \theta \le 1 \right|\] \[\cos \theta=-\frac{ 1 }{ 2 }=-\cos 60=\cos \left( 180-60 \right),\cos \left( 180+60 \right)\] \[\theta=120,240\]

  6. jmark
    • one year ago
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    putting cos theta = t, equation becomes 2t^2 + 13t = -6 = 2t^2+13t+6=0 solving (2t+1)(t+6(2t+1)(t+6)=0, t=-1/2, -6. cos theta = -1/2,-6. cos theta cannot be -6. Hence, cos theta = -1/2, theta = 2pi/3, 4pi/3

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