#2cos^2theta-4costheta-5=0#. Note that this equation looks like the quadratic equation #2x^2-4x-5=0#. So we will solve using the quadratic formula #x=(-b+- sqrt((b^2-4ac)))/(2a)#
#a=2,b=-4,c=-5#
#costheta=(4+- sqrt(16-4(2)(-5)))/4#
#costheta=(4+-sqrt56)/4#
#costheta=(4+sqrt56)/4# or #costheta=(4-sqrt56)/4#
#theta=cos^-1((4+sqrt56)/4)# or #theta=cos^-1((4-sqrt56)/4)#
#theta=#undefined or #theta=150.56# deg