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I have the following equation:

$$0 = 2e^{-x}\big(\sin(x) + 2\cos(x)\big),~x\in[0,3]$$

I must find the value of $x$ within the interval $[0,3]$.

Here is my work so far:

\begin{align} 0 &= 2e^{-x}\big(\sin(x) + 2\cos(x)\big)\newline &= \sin(x) + 2\cos(x) && \gets\text{factored out }2e^{-x}\text{ because }e^{-x}\overset{!}{\neq}0\text{ …}\newline 2\cos(x) &= -\sin(x)\newline 2\cos(x) &= -\sqrt{1 - \cos^2(x)} && \gets\text{given the Pythagorean trig identities…}\newline 4\cos^2(x) &= 1 - \cos^2(x)\newline 5\cos^2(x) &= 1\newline \cos^2(x) &= \frac{1}{5}\newline \cos(x) &= \pm\sqrt{\frac{1}{5}}\newline x &= \cos^{-1}\left(\pm\sqrt{\frac{1}{5}}\right)\text{, }x\in[0,3]\newline &\approx 1.11,\text{ }2.03 \end{align}

The real answer, as I could confirm with an online graphing calculator, is $x = 2.03$. However, $\cos^{-1}\left(+\sqrt{\frac{1}{5}}\right)$ also equals $\approx1.11$, although that solution seems to not be real, according to the graphing calculator.

Additionally, the answer key for the homework I am working on makes it obvious that $2.03$ is the only answer. (This is part of a bigger problem, but plugging $2.03$ into the next step results in what the answer key wants, and $1.11$ does not work.)

My question is as follows: How do I confirm algebraically that $1.11$ is not a real solution? If it weren't for any of the external sources that I used, I couldn't figure this out by myself, as far as my knowledge/intuition goes (as of now).

Note. In the above equation environment, I am using $\overset!\neq$ as "cannot equal".

2'5 9'2
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GPWR
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3 Answers3

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If I had to guess, it is because of "information loss" when you squared both sides in the step $$ 2\cos(x) = -\sqrt{1 - \cos^2(x)} \;\;\;\xrightarrow{\;\;\;\text{square}\;\;\;}\;\;\; 4\cos^2(x) = 1 - \cos^2(x) $$ Suppose that $\cos(x)$ were positive; then the first equality wouldn't hold ($-\sqrt x \le 0$), but the second one still could.

One notes that $\arccos(\sqrt{1/5})$ gives $2/\sqrt{5}$ on the first expression, and $-2/\sqrt 5$ on the second expression, confirming my suspicions.

In case you're confused about this "information loss", a simpler example occurs with solving $x=1$. While trivial (the solution is obvious), if we for some reason squared both sides, we get $x^2 = 1$, which now has $x=-1$ as an extraneous solution.

In general, if you apply an operation to both sides which is not "injective" (which means it can send different values to the same value, just as how $f(x) = x^2$ has $f(-1) = f(1)$ even though $-1 \ne 1$), this phenomenon can happen.

Of course, if one merely wishes to confirm that $\arccos(\sqrt{1/5})$ is not a real solution to the equation (in the sense of "real" meaning a "true" solution), one need only plug it in into the original equation they wished to solve.

For the original equation, Wolfram returns the value $$ \frac{8}{\sqrt 5} e^{-\arccos(1/\sqrt 5)} \approx 1.18 $$ which is decisively nonzero. However, $\arccos(-1/\sqrt 5)$ does return zero (Wolfram).

PrincessEev
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As an alternative

$$\sin(x) + 2\cos(x) =\sqrt 5 \sin\left(x+\arctan 2\right)=0$$

which requires

$$x=-\arctan 2 +\pi\approx 2.03$$

which indeed, as suggested in the comments, corresponds to the solution we can obtain by the simpler way (for $\cos x \neq 0$)

$$\sin(x) + 2\cos(x)=0 \iff \tan x = -2$$

By these methods we avoid possible extraneous solutions introduced by squaring the original equation.

user
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You can simply test if $\arccos(\sqrt{1/5})$ is a solution directly:

$$\begin{align} 0&\stackrel{?}{=}\sin(\arccos(\sqrt{1/5}))+2\cos(\arccos(\sqrt{1/5}))\\ 0&\stackrel{?}{=}\frac{\sqrt{5-1}}{\sqrt{5}}+2\sqrt{\frac15}\\ 0&\stackrel{\text{no}}{=}\frac{4}{\sqrt{5}} \end{align} $$

That is how you can establish that potential solution is in fact extraneous without using any calculator.

And you can likewise verify that $\arccos(-\sqrt{1/5})$ works.

2'5 9'2
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