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

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