rownania

[Bartosz89M]

rozwiąż równania

a) \(\displaystyle{ ctgx+tgx= \frac{4 \sqrt{3} }{3}}\)

b) \(\displaystyle{ tgx+ctgx = 4sin2x}\)

c) \(\displaystyle{ (cosx-sinx) ^{2} + tgx = 2sin ^{2}x}\)

d) \(\displaystyle{ ctgx+ \frac{sinx}{1+cosx}=2}\)

[kujdak]

a)
\(\displaystyle{ \frac{sin^{2}x+cos^{2}x}{sinxcosx}=\frac{4\sqrt 3}{3}\\
\frac{1}{sinxcosx}=\frac{4\sqrt 3}{3}\\
4\sqrt 3 sinxcosx=3\\
\frac{1}{2}sin2x=\frac{3}{4\sqrt 3}\\
sin2x=\frac{\sqrt 3}{2}}\)

teraz juz łatwo.

[ Dodano: 28 Marca 2008, 22:00 ]
b)
\(\displaystyle{ \frac{sin^{2}x+cos^{2}x}{sinxcosx}=4sin2x\\
\frac{1}{sinxcosx}=4sin2x\\
8sin^{2}xcos^{2}x=1\\
sinxcosx=\frac{\sqrt 2}{4}\\
\frac{1}{2}sin2x=\frac{\sqrt 2}{4}\\
sin2x=\frac{\sqrt 2}{2}}\)

[Dumel]

b) \(\displaystyle{ tgx+ctgx = 4sin2x}\)

\(\displaystyle{ \frac{sinx}{cosx}+ \frac{cosx}{sinx}=8sinxcosx}\)
\(\displaystyle{ sin^2x+cos^2x=8sin^2xcos^2x}\)
\(\displaystyle{ \frac{1}{2}=sin^22x}\)
\(\displaystyle{ |sin2x| = \frac{ \sqrt{2} }{2}}\)
\(\displaystyle{ |x|= \frac{\pi}{8}}\)

[kujdak]

c)
https://matematyka.pl/61870.htm

[ Dodano: 28 Marca 2008, 22:22 ]
d)

\(\displaystyle{ \frac{cosx}{sinx}+\frac{sinx}{1+cosx}=2\\
\frac{cosx+cos^{2}x+sin^{2}x}{sinx+sinxcosx}=2\\
\frac{1}{sinx}=2\\
sinx=\frac{1}{2}}\)