- #1

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## Homework Statement

[tex] \tan(2x)-\cot(3x)=0 [/tex], where, [tex] 0 \le x < 2\pi [/tex]

I have tried to plot the figure of this equation, but it seems so complicated.

## Homework Equations

## The Attempt at a Solution

Using the double-angle formula [tex]\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}[/tex] and

identity [tex]\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}[/tex], we get

[tex]\tan(3x)=\tan(x+2x)=\frac{\tan(x)+\tan(2x)}{1-\tan(x)\tan(2x)}[/tex],

and then

[tex]\cot(3x)=\frac{1}{\tan(x+2x)}=\frac{1-\tan(x)\tan(2x)}{\tan(x)+\tan(2x)}=\frac{1-\tan(x)\frac{2\tan(x)}{1-\tan^2(x)}}{\tan(x)+\frac{2\tan(x)}{1-\tan^2(x)}}[/tex]

After some simplification, we have

[tex]\cot(3x)=\frac{1-3\tan^2(x)}{3\tan(x)-\tan^3(x)}[/tex]

Adding our results [tex]\tan(2x)[/tex] and [tex]\cot(3x)[/tex],

the equation can be rewritten as:

[tex]5\tan(x)^4-10\tan(x)^2+1=0[/tex]

Therefore I get:

[tex]\tan(x)=\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}[/tex]

and

[tex]x=\pm\arctan\left(\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right)[/tex]

But it seems that my solution is not completed, some solutions are missing.

As Mentallic mentioned, it is not appropriate to drag out the plus minus sign in the arctan function, so the solutions should be written as

[tex]x=\arctan\left(\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right),\pi+\arctan\left(\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right)[/tex]

However, we find that two of the eight solutions are negative, where the problem requires [tex]0 \le x < 2\pi[/tex].

These two solutions are [tex]\arctan\left(-\sqrt{1+\frac{2}{5}\sqrt{5}}\right)[/tex] and [tex]\arctan\left(-\sqrt{1-\frac{2}{5}\sqrt{5}}\right)[/tex]

Since for a given value [tex]y=\tan(x)[/tex], it should correspond to two angle [tex]x[/tex] for a given range [tex]0 \le x < 2\pi[/tex]. In this case, we simply add [tex]2\pi[/tex] to those two negative solutions, the correct solutions can be obtained.

Finally, we get the results:

[tex]x=\arctan\left(\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right),\pi+\arctan\left(\pm\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right),2\pi+\arctan\left(-\sqrt{1\pm\frac{2}{5}\sqrt{5}}\right)[/tex]

However, as indicated by Mentallic at #4, we are still missing two solutions due to the fact that we simplified the equation by multiplying [tex](1-\tan^2(x))(3\tan(x)-\tan^3(x))[/tex] which is equivalent to divide the equation by [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}[/tex]. In such case it implies that we assumed [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}\neq 0[/tex] that in turn we lose the solutions of [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}=0[/tex]. As a result, we need to solve equation [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}=0[/tex] to find out the missing solutions.

Equation [tex]\frac{1}{(1-\tan^2(x))(3\tan(x)-\tan^3(x))}=0[/tex] can be rewritten as

[tex]\frac{1}{3\tan(x)-4\tan^3(x)+\tan^5(x)}=0[/tex]

That means the denominator is undefined. For [tex]\tan(x),\tan^3(x),\tan^5(x)[/tex], they have the same undefined points, which are [tex]\frac{\pi}{2},\frac{3\pi}{2}[/tex], therefore their combination [tex]3\tan(x)-4\tan^3(x)+\tan^5(x)[/tex]. In conclusion, the two missing solutions are [tex]x=\frac{\pi}{2},\frac{3\pi}{2}[/tex].

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