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)#