Question

How do i solve this equation over the interval [0degrees,360)



cot(theta)+2csc(theta)=5

Answer 1

change it to

cosØ/sinØ + 2/sinØ = 5
multiply by sinØ
cosØ + 2 = 5sinØ
square both sides
cos^2 Ø + 4cosØ + 4 = 25sin^2 Ø = 25(1 - cos^2 Ø)
cos^2Ø + 4cosØ + 4 = 25 - 25cos^2Ø
26cos^2Ø + 4cosØ - 21 = 0
cosØ = (-4 ± √2200)/52
= .82508 or -.978926

if cosØ = .82508, then Ø = 34.4° or 325.6°
if cosØ =-.978926, Ø = 168.2° or 191.8°

BUT... since we squared our equation all answers have to be verified in the original equation
x = 34.4° , LS = 5.0005 , good
x = 325.6 , LS = -3.5 ≠ RS , no good
x = 168.2 , LS = 4.999 , good
x = 191.8 , LS = -5 , no good

so x = 34.4° or x = 168.2°