We know that
1 + cot^2 Ø = csc^2 Ø
so our equation is
2(csc^ x - 1) + 3csc x = 0
2 csc^2 x + 3csc x - 2 = 0
(2cscx -1)(cscx + 2) = 0
cscx = 1/2 or cscx = -2
or
sinx = 2 (not possible) or sinx = -1/2
so x must be in III or IV
x = 7π/6 or 11π/6
find the exact zeros of the function in the interval [0, 2π)
2 cot squared x + 3 csc x=0
1 answer