How do you find the general solutions for #cot^2 x +csc x = 1#?

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How do you find the general solutions for #cot^2 x +csc x = 1#?

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Answer 1 · 2015-08-10T12:20:26

#x = n pi - pi/6(-1)^n " or " x = 2npi + pi/2 ", " n in ZZ #

Explanation:

#sin^2 x + cos^2 x = 1#
Divide by #sin^2 x# gives: #1 + cot^2 x = csc^2 x#

Therefore: #csc^2 x + csc x - 2 = 0#
#(csc x + 2)(csc x - 1) = 0#
#1/sin x = -2 " or " 1/sin x = 1#
#sin x = -1/2 " or " sin x = 1#
#x = n pi - pi/6(-1)^n " or " x = 2npi + pi/2 ", " n in ZZ #

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How do you find the general solutions for #cot^2 x +csc x = 1#?

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