How do you find the solution to $csc^2theta-6csctheta+8=0$ if $0<=theta<2pi$ ?

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Answer 1 · 2016-12-19T05:46:13

$theta=pi/6 or theta=5pi/6 or theta=arcsin(1/4) or theta=pi-arcsin(1/4)$

You can use the quadratic formula, in the form:

$x=-b/2+-sqrt(( b/2)^2-c)$ ,

useful when b is even and $a=1$ .

Then

$csctheta=3+-sqrt(9-8)$

$csctheta=2 or csctheta=4$

that's

$sintheta=1/2 or sintheta=1/4$

$theta=pi/6 or theta=(5pi)/6 or theta=arcsin(1/4) or theta=pi-arcsin(1/4)$

How do you find the solution to csc^2theta-6csctheta+8=0csc^2theta-6csctheta+8=0 if 0<=theta<2pi0<=theta<2pi?