Rozwiąż równania

[t5]

\(\displaystyle{ tgx - 1= sinx - cosx}\)

\(\displaystyle{ tgx + ctgx=2 \sqrt{2}}\)

\(\displaystyle{ tg ^{3}x - 3tgx= tg ^{2}x -3}\)

[Nakahed90]

\(\displaystyle{ \tg^{3}x-\tg^{2}x-3\tg x+3=0}\)
\(\displaystyle{ \tg^{2}x(\tg x -1)-3(\tg x -1)=0}\)
\(\displaystyle{ (\tg x -1)(\tg^{2} x-3)=0}\)
\(\displaystyle{ (\tg x -1)(\tg x-\sqrt{3})(\tg+\sqrt{3})=0}\)
\(\displaystyle{ \tg x=1 \iff x=\frac{\pi}{4}+k \pi k\in C}\)
\(\displaystyle{ \tg x=\sqrt{3} \iffx=\frac{\pi}{3}+k \pi k\in C}\)
\(\displaystyle{ \tg x=-\sqrt{3} \iffx=\frac{-\pi}{3}+k \pi k\in C}\)

\(\displaystyle{ \frac{\sin x}{\cos x}-1=\sin x-\cos x}\)
\(\displaystyle{ \frac{\sin x-\cos x}{\cos x}=\sin x-\cos x}\)
\(\displaystyle{ \cos x=1 \iff x=2k \pi k\in C}\)

[frej]

Nakahed90, musisz jeszcze rozważyć przypadek \(\displaystyle{ sin{x}-cos{x}=0}\), chyba nie muszę mówić dlaczego

[leszek1820]

\(\displaystyle{ \frac{sinx}{cox}-1=sinx-cox /*cos\\
sinx-cox=sinxcox- cos ^{2}x \\
cos ^{2}x- sinxcox-cos+sinx=0\\
cosx(cox-sinx)-1(cos-sinx)=0\\
(cosx-sinx)(cos-1)=0\\
cos-sinx=0\\
cosx=1}\)
dalej juz proste chyba sobie poradzisz

drugie
\(\displaystyle{ \frac{1}{sinxcox}= 2 \sqrt{2} \\
1=2 \sqrt{2} sinxcox\\
\sqrt{2} /2=\sin 2x\text{ bo }2\sin x \cos x = \sin 2x}\)
i dalej juz z gorki