Question

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

Answer 1

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

Answer 2

If like ab854 asked, hint: quadratic equation.

Answer 3

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

Answer 4

thank you guys.

Answer 5

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

Answer 6

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