Suppose I have a few ecuations:
$$\cos^2(x) = \sin(x)\cos(x) \Rightarrow cos(x) = sin(x) $$
$$ x^2 + 3x \ge 2x \Rightarrow x(x+3) \ge 2x \Rightarrow x+3 \ge 2$$
Which of them are true and why?
Basically, when can one simplify an equation or an inequation with common terms that contain x (a variable) that we need to solve for? We may lose solutions.
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$\begingroup$For equations one can always apply a function on both sides, this includes dividing(unless of course you divide by $0$), squaring, subtracting etc. This is because of the obvious but nevertheless useful to note fact that $x = y$ implies $f(x) = f(y)$.
For inequalities this is not the case, because essentially $x \leq y$ does not imply $f(x) \leq f(y)$ for all functions, but there are a lot of functions for which it is known whether the function is decreasing or increasing. For example, multiplying by some $c>0$ is an increasing function, in other words: $x \leq y$ implies $cx \leq cy$.
Now we can apply the preceding discussion to your examples. If $\cos^2(x) = \sin(x) \cos(x)$, we can divide both sides by $\cos(x)$, obtaining $\cos(x) = \sin(x)$, if $\cos(x) \neq 0$. In the case $\cos(x) = 0$, the equation does not imply anything new. Now for your inequality, only the last step is dubious, because in the case $x<0$, $1/x< 0$, so multiplying both sides by $1/x$ turns around the sign and we get $x+3 \leq 2$. If $x>0$ the last step holds, but then this implies $x \geq -1$ which does not give us any new information, since $x> 0$. However, adding and subtracting is increasing, so $x^2 + 3x \geq 2x$ if and only if $x^2+x \geq 0$. We see this equation is satisfied if $x \geq 0$. if $x \leq 0$, we have $x \leq -1$. Conversely if $x \leq -1$ we have $x+3 \leq 2$, and multiplying both sides by $x$ and noticing $x<0$ we have $x(x+3) \geq 2x$. So the solutions of the inequality are all $x$ for which $x \geq 0$ and $x \leq -1$.
Finally a general tip for this kind of confusion: always remind yourself of what it is you want to do. Do you want to find $x$ such that $f(x) = g(x)$? Do you want to validate whether $f(x) = g(x)$ for all $x$? Do you want to see wheter $f(x) = g(x)$ implies some contradiction? Keep using logic, if you take a step, think about whether it is reversible (in other words, whether it is 'if and only if'), and always remember that an equation without an explanation means nothing.
$\endgroup$ 3 $\begingroup$If $f(x)$ is monotonic then you can simplify as if the equation is an equality. Monotonic means the first derivative never changes sign. Broadly speaking this means the function always goes up with $x$ or down with $x$ and never changes.
In your first example, $cos^2(x)$ isn't strictly $1:1$ because in some places it goes up with $x$ and in other paces it goes down.
In your 2nd, the quadratic form will again go both up and down due to positive and negative square roots.
As a rule, find the source of the duotonicity and this will lead you to find the multiple solutions. In the case of the quadratic you will need to allow for the positive and negative square roots. In the case of the trigonometric identities you need to account for the cyclicity of the functions with period $2\pi$.
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