团圆之际，千万别忘了那些至亲至爱的人-网通

Question

I'm having trouble solving trigonometric equations in general for Pre-Calculus. I often get different answers for the same problem depending on the method I use.

This also happened to me during Algebra 2 when solving radical equations, but this was pretty straightforward since I just had to double-check the answers by substituting to the original equation.

For example, when solving $$ \sqrt{x + 2} = x - 4 $$

you get the solution x = 2 or 7, but you know that x = 2 is an extraneous solution when you plug it in. $$$$

However, when solving trigonometric equations such as: $$\tan(2x) - \cot(x) = 0$$ for $x \in [0, 2\pi)$,

When using method 1 of $\frac{2 \tan x}{1 - \tan^2 x} - \frac{1}{\tan x} = \frac{3 \tan^2 x - 1}{\tan x (1 - \tan^2 x)} = 0$ using the double angle formula, I get $\tan x = \pm \frac{1}{\sqrt{3}}$
, which results in $$x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$$.

But when using method 2 of $\tan(2x) = \cot(x)$,

$\cot(x) = \tan\left(\frac{\pi}{2} - x\right)$, so $\tan(2x) = \tan\left(\frac{\pi}{2} - x\right)$
and $2x = \frac{\pi}{2} - x + k\pi$ where $k \in \mathbb{Z}$
$x = \frac{\pi}{6} + \frac{k\pi}{3}$.

Therefore, $$x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$$

See how there are two new solutions for method 2 that is valid when plugging in to the original question? I really have no idea why different methods result in different solutions.

Can anybody please explain why this happens?

Answer 1

One of the reasons is that when you transform an equation in certain ways, you do not get an equivalent equation but rather an equation that includes the solutions of the original one. Consider, for example, the equation $x=\sqrt2$. Obviously there is only one solution. Now square it: $x^2=2$. There are now two solutions. The lesson is that when one transforms equations, one can't do it mindlessly, but needs rather to keep track of what happens with solutions.

As an exercise, try to see how this applies to your trigonometric example.

Answer 2

The discrepancy comes from the double angle formula:

$$\tan(2x)=\dfrac{2\tan(x)}{1-\tan^2(x)}$$ does not work for $x=\dfrac\pi2$.

Answer 3

When dividing through or multiplying through, one should check that one has not divided through or multiplied through by zero.

Why multiplying through by zero is unhelpful: "$x = 2$" becomes "$0 = 0$" leading one to the false conclusion that the original equation is a tautology and $x$ can be anything. Sometimes better to explain as "multiplying through by zero has destroyed all information in this equation, so it is no longer of any use." The seems "silly" when written as the above, but you are starting to work in expressions that are zero in less obvious ways. For instance multiplying $\theta = \pi$ through by $\tan \theta$ yields $0 = 0$.

Dividing by zero is just as unhelpful since it involves an undefined operation. The correct inference adds a clause to check for division by zero. The $\tan(2x)$ is a division by zero when $x \in \{\pi/2, 3\pi/2\}$ (restricting to $x \in [0, 2\pi)$). When we use $$ \tan(2x) = \frac{2 \tan x}{1-\tan^2 x} $$ we typically skim over the fact that both sides involve division by zero when $x \in \{k \pi + \pi/2 : k \in \Bbb{Z}\}$ because both sides are undefined at the same time and identities are only required to hold on the common domain of the functions appearing on the left and the right. Equations expect a little more, so the correct inference is $$ \tan(2x) = \frac{2 \tan x}{1-\tan^2 x} \quad\text{ or }\quad 1 - \tan^2 x = 0 $$ and that second clause provides the solutions you are missing. One can also use "$\tan(2x) \text{ is undefined}$" or "$\cos(2x) = 0$" (since $\tan u = \sin u/\cos u$ explicitly exhibits the division by zero that is causing this extra clause to appear).

Answer 4

You've been lied to about how solving equations works.

Mathematics it taught as a series of patterns and techniques; that is useful for someone doing computation by hand, but it what you are taught is basically full of lies.

When you take $$\sqrt{x} = 2$$ and you "solve" it by tasquaring both sides $$x=4$$, what you are actually doing is claiming that the two equations are equivalent. But many of the steps you take don't actually involve equivalent equations; they are equivalent almost always, or the first implies the second, or the second implies the first.

Here, the same process starting with $-\sqrt{x} = 2$ where you square both sides to $x=4$ gives a spurious solution; the problem is $a=b$ implies $\sqrt{a}=\sqrt{b}$ but not the other way around.

Doing such transformations can lead you to an answer, but can also include extra "solutions" which are not valid; in fact, you might end up with none of the solutions being valid (if the original equation had no answers, for example).

In simple cases, this is often when you implicitly multiply by 0 (clearing denominators, or some trig identities), but not always.

To do mathematics more carefully, you need to understand which statement implies the other and in which cases this implication is valid (and when it is invalid).

If your goal is finding a solution, you want backwards implication - if you start with equation A, you want the next step B to IMPLY A. Then any solution you find will solve A.

If your goal is showing there is no solution, then you want forward implication. If you start with A, the next step B should be IMPLIED by A. Then if you come up with a contradiction you know there is no solution to A.

If your goal is to find every solution, then you want bi-directional implication - equivalence - in each step.

If you have limited connection - the connection isn't valid for certain values - you have to "fork" your solution (do the rest of it twice) and check that case separately. Sometimes that is easy; you can just inspect that case. In your case, you aren't doing that, and those "forks" are showing up in your simplified equation and you can't tell they are invalid solutions.

When taught to manipulate equations in practice people skip this detail, or don't focus on it very heavily, or fix it in a post-processing pass where they say "and then check your solutions to make sure they work". This is a practical way to handle the problem, but it means you'll be doing unsound work in the middle and might not even realize it.

You are seeing the effects of that unsound work. Now quite often, especially with toy problems you get for education purposes, the unsoundness is just a quirk.

Answer 5

When rewriting expressions you should always consider when solutions are still valid. Plugging your set of solutions of the 2nd method will yield a problem for the extraneous solutiotions, i.e. $x=\pi/2$ and $x=3\pi/2$. Plugging those values in for $\tan(2x)$ gets you an indeterminate answer.

So you have the techniques down, but like in your radical equation, you need to check if solutions are valid.

Answer 6

Your first solution is incomplete:

$$\frac{3 \tan^2{x}-1}{\tan x \cdot (1-\tan^2 x)}=0$$

Values like $x=\frac{\pi}{2}+n\pi$ are also a solution of that equation (extremely simplified: the tangent goes to infinity and dividing by infinity goes to zero).