What is x in Trig Problem with given constraints?

[Greg]

Find $$x$$ such that $$\tan x=1+\sqrt2$$ and $$0<x<\dfrac{\pi}{2}$$.

 

[Euge]

Spoiler

Since $1 + \tan x = 2 + \sqrt{2}$ and $1 - \tan x = -\sqrt{2}$, then

$$\tan\left(\frac{\pi}{4}-x\right) = \frac{1 - \tan x}{1 + \tan x} = \frac{-\sqrt{2}}{2 + \sqrt{2}} = -\frac{1}{\sqrt{2}+1}= -\frac{1}{\tan x}.$$

Therefore,

$$1 + \tan x\, \tan\left(\frac{\pi}{4}-x\right) = 0,$$

which implies

$$\frac{1 + \tan x\tan\left(\frac{\pi}{4}-x\right)}{\tan x - \tan\left(\frac{\pi}{4}-x\right)} = 0,$$

or,

$$\cot\left(x + \left(\frac{\pi}{4}-x\right)\right) = 0,$$

that is,

$$\cot\left(2x - \frac{\pi}{4}\right) = 0.$$

One of the solutions to this equation is such that $2x - \frac{\pi}{4} = \frac{\pi}{2}$, or $x = \frac{3\pi}{8}$. So all solutions are of the form $\frac{3\pi}{8} + n\pi$, where $n$ ranges over the integers. Since we seek those solutions in $(0, \frac{\pi}{2})$, the only root of the original equation is $x = \frac{3\pi}{8}$.

 

[kaliprasad]

Spoiler

$tan x = 1+ \sqrt2$
we need to get rid of $\sqrt2$
so $\tan x -1 = \sqrt 2$
or $\tan^2 x-2\tan x + 1 = 2$
or $\tan ^2 x -1 = 2\tan x$
or $\tan 2x = -\dfrac{2\tan x}{\tan^2x -1} = -1 = \tan (n\pi + \dfrac{3\pi}{4})$
$x = \dfrac{1}{2} (n\pi + \dfrac{3\pi}{4})$
so x in range is $\frac{3\pi}{8}$

 

[anemone]

Find $$x$$ such that $$\tan x=1+\sqrt2$$ and $$0<x<\dfrac{\pi}{2}$$.

My solution:

Spoiler

$\tan x=1+\sqrt2$

$\sin x=\cos x+\sqrt2 \cos x$

$\sin x-\cos x=\sqrt2 \cos x$

$\sqrt2(\sin (x-45^\circ))=\sqrt2 \sin(90^\circ- x)$

$\therefore 2x=135^\circ\,\,\,\rightarrow\,\,\,x=67.5^\circ$ for $$0<x<90^\circ$$.

 

[Greg]

Find $$x$$ such that $$\tan x=1+\sqrt2$$ and $$0<x<\dfrac{\pi}{2}$$.

Spoiler

Construct right-angled triangle $$ABC$$ such that $$\angle{ABC}=90^\circ,\overline{AB}=1,\overline{BC}=1+\sqrt2$$. Construct point $$D$$ on $$\overline{BC}$$ such that $$\overline{BD}=1$$.
Then $$\angle{CAD}=22.5^\circ,\angle{BAD}=45^\circ,\angle{BAC}=\angle{CAD}+\angle{BAD}=67.5^\circ$$ or $$\dfrac{3\pi}{8}$$ radians.