rozwiąż równanie:
sin3x+cos3x=\(\displaystyle{ \sqrt{2}}\)
sinx+\(\displaystyle{ \sqrt{3}}\)cosx-1=0
równanie
- Tristan
- Użytkownik
- Posty: 2353
- Rejestracja: 24 kwie 2005, o 14:28
- Płeć: Mężczyzna
- Podziękował: 27 razy
- Pomógł: 557 razy
równanie
1) \(\displaystyle{ sin 3x + cos 3x= \sqrt{2}}\)
zał:\(\displaystyle{ z=3x}\)
\(\displaystyle{ sin z + cos z= \sqrt{2}}\)
\(\displaystyle{ \sqrt{2} cos( z- \frac{\pi}{4})= \sqrt{2}}\)
\(\displaystyle{ cos(z - \frac{\pi}{4})=1}\)
\(\displaystyle{ z- \frac{\pi}{4}=2k \pi , k C}\)
\(\displaystyle{ 3x=2k \pi + \frac{\pi}{4}, k C}\)
\(\displaystyle{ x= \frac{\pi}{12} + 2k \pi, k C}\)
2) \(\displaystyle{ sin x + \sqrt{3} cos x -1=0}\)
\(\displaystyle{ 2( \frac{1}{2} sin x + \frac{ \sqrt{3}}{2} cos x)=1}\)
\(\displaystyle{ 2( cos x cos 30^{\circ}+ sin x sin 30^{\circ})=1}\)
\(\displaystyle{ 2 cos(x-30^{\circ})=1}\)
\(\displaystyle{ cos(x- \frac{ \pi}{6})=\frac{1}{2}}\)
\(\displaystyle{ x-\frac{ \pi}{6}= \frac{\pi}{3}+2k \pi x-\frac{ \pi}{6}=-\frac{\pi}{3} + 2k \pi, k C}\)
\(\displaystyle{ x=\frac{\pi}{2}+2k \pi x=-\frac{ \pi}{6}+2k \pi, k C}\)
zał:\(\displaystyle{ z=3x}\)
\(\displaystyle{ sin z + cos z= \sqrt{2}}\)
\(\displaystyle{ \sqrt{2} cos( z- \frac{\pi}{4})= \sqrt{2}}\)
\(\displaystyle{ cos(z - \frac{\pi}{4})=1}\)
\(\displaystyle{ z- \frac{\pi}{4}=2k \pi , k C}\)
\(\displaystyle{ 3x=2k \pi + \frac{\pi}{4}, k C}\)
\(\displaystyle{ x= \frac{\pi}{12} + 2k \pi, k C}\)
2) \(\displaystyle{ sin x + \sqrt{3} cos x -1=0}\)
\(\displaystyle{ 2( \frac{1}{2} sin x + \frac{ \sqrt{3}}{2} cos x)=1}\)
\(\displaystyle{ 2( cos x cos 30^{\circ}+ sin x sin 30^{\circ})=1}\)
\(\displaystyle{ 2 cos(x-30^{\circ})=1}\)
\(\displaystyle{ cos(x- \frac{ \pi}{6})=\frac{1}{2}}\)
\(\displaystyle{ x-\frac{ \pi}{6}= \frac{\pi}{3}+2k \pi x-\frac{ \pi}{6}=-\frac{\pi}{3} + 2k \pi, k C}\)
\(\displaystyle{ x=\frac{\pi}{2}+2k \pi x=-\frac{ \pi}{6}+2k \pi, k C}\)