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

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1. anonymous

$2\cos^2(\theta) + 13\cos(\theta) = -6$ Like that?

2. barrycarter

3. anonymous

yes like that, how did you do the theta sign?

4. anonymous

thank you guys.

5. anonymous

$2 \cos ^2\theta+13\cos \theta+6=0$ $2 \cos ^2\theta+12\cos \theta+\cos \theta+6=0$ $2\cos \theta \left( \cos \theta+6 \right)+1\left( \cos \theta+6 \right)=0$ $\left( \cos \theta+6 \right)\left( 2\cos \theta+1 \right)=0$ $Either~\cos \theta=-6 ~(rejected),~as \left| \cos \theta \le 1 \right|$ $\cos \theta=-\frac{ 1 }{ 2 }=-\cos 60=\cos \left( 180-60 \right),\cos \left( 180+60 \right)$ $\theta=120,240$

6. anonymous

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