Równanie trygonometryczne

bykoxD · Post autor: **bykoxD** » 3 mar 2009, o 17:42

\(\displaystyle{ sin2x + cosx = 0}\)

Nakahed90 · Post autor: **Nakahed90** » 3 mar 2009, o 17:57

\(\displaystyle{ 3sinx=2(1-sin^{2}x)=2-2sin^{2}x}\)
\(\displaystyle{ 2sin^{2}x+3sinx}-2=0}\)
\(\displaystyle{ 2sin^{2}x-sinx+4sinx-2=0}\)
\(\displaystyle{ sinx(2sinx-1)+2(2sinx-1)=0}\)
\(\displaystyle{ (2sinx-1)(sinx+2)=0}\)
\(\displaystyle{ sinx=\frac{1}{2} \vee sinx=-2\not\in <-1,1>}\)

bykoxD · Post autor: **bykoxD** » 3 mar 2009, o 18:08

dzieki

Nakahed90 · Post autor: **Nakahed90** » 3 mar 2009, o 19:04

\(\displaystyle{ sin^{2}3x=1}\)
\(\displaystyle{ sin3x=1 \vee sin3x=-1}\)
\(\displaystyle{ 3x=\frac{\pi}{2}+2k\pi \vee 3x=\frac{3\pi}{2}+2k\pi k\in C}\)
\(\displaystyle{ x=\frac{\pi}{6}+k\frac{2\pi}{3} \vee x=\frac{\pi}{2}+k\frac{2\pi}{3} \wedge k\in C}\)

\(\displaystyle{ 2cos^{2}2x=1}\)
\(\displaystyle{ cos^{2}2x=\frac{1}{2}}\)
\(\displaystyle{ cos2x=\frac{\sqrt{2}}{2} \vee cos2x=-\frac{\sqrt{2}}{2}}\)
\(\displaystyle{ 2x= \pm \frac{\pi}{4}+2k\pi \vee 2x= \pm \frac{3\pi}{4}+2k\pi \wedge k\in C}\)
\(\displaystyle{ x =\pm \frac{\pi}{8}+k\pi \vee x= \pm \frac{3\pi}{8}+k\pi \wedge k\in C}\)

bykoxD · Post autor: **bykoxD** » 3 mar 2009, o 19:42

\(\displaystyle{ cos2x= \frac{ \sqrt{2}}{2} \vee cos2x=- \frac{ \sqrt{2}}{2}}\)
skad to sie bierze?

Nakahed90 · Post autor: **Nakahed90** » 3 mar 2009, o 19:46

Wyciągnąłem obustronnie pierwiastek.

bykoxD · Post autor: **bykoxD** » 3 mar 2009, o 19:48

nie rozumiem ;/

Nakahed90 · Post autor: **Nakahed90** » 3 mar 2009, o 19:52

\(\displaystyle{ cos^{2}2x=\frac{1}{2}|\sqrt{}}\)
\(\displaystyle{ cos2x=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2} \vee cos2x=-\sqrt{\frac{1}{2}}=-\frac{\sqrt{2}}{2}}\)

matshadow · Post autor: **matshadow** » 3 mar 2009, o 19:52

\(\displaystyle{ cos^{2}2x=\frac{1}{2}\leftrightarrow cos2x=\frac{\sqrt{2}}{2} \vee cos2x=-\frac{\sqrt{2}}{2}}\)