1. ## goniometric function

I would like to know, if there are some mistakes or absences.

Thank you so much, M.

$\displaystyle 2sin^2x+cos^2x+sinxcosx=1$

$\displaystyle 2(1-cos^2x)+cos^2x+sinxcosx=1$

$\displaystyle 2-2cos^2x+cos^2x+sinxcosx=sin^2x+cos^2x$

$\displaystyle 2-cos^2x+sinxcosx=sin^2x+1-sin^2x$

$\displaystyle 2-(1-sin^2x)+sinxcosx=1$

$\displaystyle 2-1+sin^2x+sinxcosx=1$

$\displaystyle 1+sin^2(x)+sinxcosx=1$

$\displaystyle sin^2x+sinxcosx+0$

$\displaystyle sinx(sinx+cosx)=0$

1. CASE

$\displaystyle sinx=0$

$\displaystyle x=k*pi$

2. CASE

$\displaystyle sinx+cosx=0$

$\displaystyle \frac{sinx}{cosx}+1=0$

$\displaystyle \frac{sinx}{cosx}=-1$

$\displaystyle tanx=-1$

$\displaystyle x_1=pi/4+k*pi$
$\displaystyle x_2=3/4pi+k*pi$

solved thx

3. ## Re: goniometric function

Hello, majovski!

There's a small error at the end: $\displaystyle x_1 \,=\,-\tfrac{\pi}{4} + k\pi$

$\displaystyle 2\sin^2\!x+\cos^2\!x+\sin x\cos x\:=\:1$

I would solve it like this . . .

$\displaystyle \sin^2\!x + \sin^2\!x + \cos^2x + \sin x\cos x \;=\;1$

$\displaystyle \sin^2\!x + 1 + \sin x\cos x \;=\;1$

$\displaystyle \sin^2\!x + \sin x\cos x \;=\;0$

$\displaystyle \sin x(\sin x + \cos x) \;=\;0$

$\displaystyle [1]\;\sin x \,=\,0 \quad\Rightarrow\quad x \,=\,k\pi$

$\displaystyle [2]\;\sin x\,+\cos x \,=\,0 \quad\Rightarrow\quad \tan x \,=\,-1 \quad\Rightarrow\quad x \,=\,-\tfrac{\pi}{4} + k\pi$

4. ## Re: goniometric function

hello and thanks a lot :-)

,

,

### gionometric function

Click on a term to search for related topics.