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I was working on $\frac{1+\cos x}{\sin x}=-1$. Solving for $x$ algebraically gave me three solutions $x=\frac{\pi}{2},\pi,\frac{3\pi}{2}$. However, graphically, as well as substituting these solutions to the original equation, only $x=\frac{3\pi}{2}$ is an actual solution. I realized that the other false solutions show up when i square both sides after multiplying by $\sin x$ on both sides.

So when would $f(x)=g(x)$ and $f(x)^2=g(x)^2$ have the same solution sets where I would not have to worry about extraneous solutions?

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    The two expressions are equivalent iff you know that $f(x)$ and $g(x)$ have the same sign. – 2017-01-28

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Note that $f(x)^2=g(x)^2$ for a particular point $x$ is equivalent to $|f(x)|=|g(x)|$, which is satisfied when $f(x)=g(x)$, but also when $f(x)=-g(x)$.

If you attempt to solve $f(x)=g(x)$ by solving $f(x)^2=g(x)^2$, then all the solutions of $f(x)=-g(x)$ will also be obtained. You can state hypotheses on $f$ and $g$ so that what you're saying ocurrs, (for example, if $f(x)$ and $g(x)$ have the same sign).

In your case, note that $\pi$ cannot be a solution, because the denominator of $\frac{1+\cos x}{\sin x}$ vanishes.

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    `|f(x)|=|g(x)|, which is satisfied when f(x)=g(x), but also when f(x)=−g(x)`. The condition $|f(x)|=|g(x)|$ is correct, but it's worth emphasizing that it implies $f(x)=g(x)$ or $f(x)=-g(x)$ only pointwise for individual $x$ values, *not* as functions across the entire domain. – 2017-01-28
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    Yes, that's true, and could lead to mistake. Thanks for the correction. – 2017-01-28