How do you find all solutions of the equation #(1+cosx)/(1-cosx)=0# in the interval [0,2pi)?

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Answer 1 · 2015-04-17T04:08:54

#f(x) = (1 + cos x)/(1 - cos x) = 0#

Multiply both numerator and denominator by #(1 - cos x).#

#f(x) = (1 - cos^2 x)/(1 - cos x)^2 = sin^2/(1 - cos x)^2#

Solve #sin^2 x = 0 --> x = 0 and x = pi#.

However, the function is undefined at #cos x = 1 --> x = 0# and #2pi#.

Then, there is one answer: #x = pi.#

Check:
#x = pi --> (cos x + 1) = -1 + 1 = 0; (cos x - 1) = -2 --> f(x) = 0/-2.# Correct