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anonymous
 one year ago
solve for 2cos^2theta + 13costheta = 6. theta is >= 0 and < 360 degrees
anonymous
 one year ago
solve for 2cos^2theta + 13costheta = 6. theta is >= 0 and < 360 degrees

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2\cos^2(\theta) + 13\cos(\theta) = 6\] Like that?

barrycarter
 one year ago
Best ResponseYou've already chosen the best response.0If like ab854 asked, hint: quadratic equation.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes like that, how did you do the theta sign?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2 \cos ^2\theta+13\cos \theta+6=0\] \[2 \cos ^2\theta+12\cos \theta+\cos \theta+6=0\] \[2\cos \theta \left( \cos \theta+6 \right)+1\left( \cos \theta+6 \right)=0\] \[\left( \cos \theta+6 \right)\left( 2\cos \theta+1 \right)=0\] \[Either~\cos \theta=6 ~(rejected),~as \left \cos \theta \le 1 \right\] \[\cos \theta=\frac{ 1 }{ 2 }=\cos 60=\cos \left( 18060 \right),\cos \left( 180+60 \right)\] \[\theta=120,240\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0putting 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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