2cos^2theta-4costheta-5=02cos2θ−4cosθ−5=0. Note that this equation looks like the quadratic equation 2x^2-4x-5=02x2−4x−5=0. So we will solve using the quadratic formula x=(-b+- sqrt((b^2-4ac)))/(2a)x=−b±√(b2−4ac)2a
a=2,b=-4,c=-5a=2,b=−4,c=−5
costheta=(4+- sqrt(16-4(2)(-5)))/4cosθ=4±√16−4(2)(−5)4
costheta=(4+-sqrt56)/4cosθ=4±√564
costheta=(4+sqrt56)/4cosθ=4+√564 or costheta=(4-sqrt56)/4cosθ=4−√564
theta=cos^-1((4+sqrt56)/4)θ=cos−1(4+√564) or theta=cos^-1((4-sqrt56)/4)θ=cos−1(4−√564)
theta=θ=undefined or theta=150.56θ=150.56 deg