Solve the following quadratic trigonometric equation for all values x E [0, 2π] algebraically

Question

csc2 (x) + 5csc (x) + 6 = 0

Answer 1

Let y = csc(x)

The given eq becomes : y² + 5y + 6 = 0 => (y+2)(y+3) = 0. => y = -3 or -2.

So csc(x) = -3 or -2. Find x from here.

Answer 2

csc^2(x) + 5csc(x) + 6 = 0

let z = csc(x)

then

z^2 +5z + 6 = 0

z = -5/2 + or - (1/2)sqr(25-24)) = -2.5 + .5 or -2.5 -.5 = -2 or -3

csc(x) = -2 or -3

sin(x) = 1/csc(x) = -1/2 or - 1/3

x = sin^-1(-1/2) or sin^-1(-1/3)

x = 210 or 330 degrees, or 199.47 or 340.53 degrees

x = 199.47, 210, 330, 340.53 degrees or x= 1.11pi, 7pi/6, 11pi/6, 1.89pi radians

sines and cosecants are negative only in quadrants III and IV

(just curious, if maybe the problem was really csc^2(x) -5csc(x) + 6 =0 ?? did that middle sign get copied wrong? with + instead of - ?? if so, it solves more easily as it factors,as if the problem had been set up to factor.

z^2 -5z + 6 = (z-3)(z-2) = 0, z= 2,3, csc(x) = 2 or 3, sin(x) = 1/2, 1/3, x = 30 or 150 degrees or 19.47 or 160.53 degrees, x = 19.47, 30, 150, 160.53 or x = 0.11pi, pi/6, 5pi/6, 0.89pi radians)