równania trygonometryczne

[Przemkooo]

Mam takie trzy równania:

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

2.\(\displaystyle{ (1-tgx) (1 + sin2x) = 1 + tgx}\)

3.\(\displaystyle{ sinx tgx - \sqrt[2]{3} = tgx - \sqrt[2]{3}sinx}\)

[Lady Tilly]

2)
\(\displaystyle{ cos2\alpha+sin2\alpha=1+2tg\alpha}\)
\(\displaystyle{ tg\frac{\alpha}{2}=\frac{1-cos}{sin\alpha}}\)

[sztuczne zęby]

1)

\(\displaystyle{ (cosx - sinx)^2 + tgx = 2sin^2x \\
x\neq\frac{\pi}{2}+k\pi\\
cos2x-sin2x+tgx=0\\
\frac{cosxcos2x-cosxsin2x+sinx}{cosx}=0}\)
\(\displaystyle{ cosxcos2x-cosxsin2x+sinx=0\\
cosx-sinx=0\\
sin(x-\frac{\pi}{4})=0\\
x=\frac{\pi}{4}+k\pi}\)

[Tristan]

\(\displaystyle{ \sin x tg x - \sqrt{3}=tg x - \sqrt{3} \sin x \\ \sin x tg x - tg x + \sqrt{3} \sin x - \sqrt{3}=0 \\ tg x ( \sin x -1)+ \sqrt{3} ( \sin x -1)=0 \\ ( \sin x -1)( tg x + \sqrt{3})=0 \\ \sin x =1 \tg x = -\sqrt{3}}\)
Myślę, że dalej już sobie poradzisz.