Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]sin\alpha +cos\alpha =\frac{4}{3} /^{2}[/tex]
[tex]sin^{2}\alpha +2sin\alpha cos\alpha +cos^{2}\alpha =\frac{16}{9}[/tex]
[tex]1 +2sin\alpha cos\alpha =\frac{16}{9}[/tex]
[tex]2sin\alpha cos\alpha =\frac{16}{9}-1[/tex]
[tex]2sin\alpha cos\alpha =\frac{7}{9}[/tex] /:2
[tex]sin\alpha cos\alpha =\frac{7}{18}[/tex]
[tex]sin\alpha(\frac{4}{3} - sin\alpha ) =\frac{7}{18}[/tex]
[tex]\frac{4}{3}sin\alpha -sin^{2}\alpha =\frac{7}{18}[/tex] /*(-18)
[tex]18sin^{2}\alpha -24sin\alpha +7=0[/tex]
Δ=(-24)² - 4 * 18 * 7 = 72
[tex]\sqrt{delty}=6\sqrt{2}[/tex]
[tex]sin\alpha =\frac{24-6\sqrt{2} }{2*18}=\frac{6(4-\sqrt{2} )}{36}=\frac{4-\sqrt{2} }{6}[/tex] [tex]sin\alpha =\frac{24+6\sqrt{2} }{2*18}=\frac{6(4+\sqrt{2} )}{36}=\frac{4+\sqrt{2} }{6}[/tex]
[tex]cos\alpha =\frac{4}{3}-\frac{4-\sqrt{2} }{6}=\frac{4+\sqrt{2} }{6}[/tex] [tex]cos\alpha =\frac{4}{3}-\frac{4+\sqrt{2} }{6}=\frac{4-\sqrt{2} }{6}[/tex]
