How do you solve 2cos^2theta-4costheta-5=02cos2θ4cosθ5=0 in the interval 0<=x<=2pi0x2π?

1 Answer
Oct 3, 2016

theta=150.56θ=150.56 deg

Explanation:

2cos^2theta-4costheta-5=02cos2θ4cosθ5=0. Note that this equation looks like the quadratic equation 2x^2-4x-5=02x24x5=0. So we will solve using the quadratic formula x=(-b+- sqrt((b^2-4ac)))/(2a)x=b±(b24ac)2a
a=2,b=-4,c=-5a=2,b=4,c=5

costheta=(4+- sqrt(16-4(2)(-5)))/4cosθ=4±164(2)(5)4

costheta=(4+-sqrt56)/4cosθ=4±564

costheta=(4+sqrt56)/4cosθ=4+564 or costheta=(4-sqrt56)/4cosθ=4564

theta=cos^-1((4+sqrt56)/4)θ=cos1(4+564) or theta=cos^-1((4-sqrt56)/4)θ=cos1(4564)
theta=θ=undefined or theta=150.56θ=150.56 deg