Matematyka

Proszę o rozwiązanie zadań z załacznika

Odpowiedź

DJoooo

1. a) [latex]|ctgx-1|=0\ctgx-1=0\ctgx=1\x=frac{pi}{4}+kpi[/latex] b) [latex]|2cosx|=1\2cosx=1 vee 2cosx=-1\cosx=frac{1}{2} vee cosx=-frac{1}{2}\x=frac{pi}{3}+2kpi vee x=frac{2}{3}pi+2kpi vee x=frac{4}{3}pi+2kpi vee x=frac{5}{3}pi+2kpi\x=frac{pi}{3}+kpi vee x=frac{2}{3}pi+kpi[/latex] c) [latex]|3tgx|=sqrt{3}\3tgx=sqrt{3} vee 3tgx=-sqrt{3}\tgx=frac{sqrt{3}}{3} vee tgx=-frac{sqrt{3}}{3}\x=frac{pi}{6}+kpi vee x=frac{5}{6}pi+kpi[/latex] d) [latex]|4sinx|=2\4sinx=2 vee 4sinx=-2\sinx=frac{1}{2} vee sinx=-frac{1}{2}\x=frac{pi}{6}+2kpi vee x=frac{5}{6}pi+2kpi vee x=frac{11}{6}pi+2kpi vee x=frac{7}{6}pi+2kpi\x=frac{pi}{6}+kpi vee x=frac{5}{6}pi+kpi[/latex] 2. a) [latex]ctg^2x=3\ctgx=sqrt{3} vee ctgx=-sqrt{3}\x=frac{pi}{6}+kpi vee x=frac{5}{6}pi+kpi[/latex] b) [latex]2sin^2x=1\sin^2x=frac{1}{2}\sinx=frac{sqrt{2}}{2} vee sinx=-frac{sqrt{2}}{2}\x=frac{pi}{4}+2kpi vee x=frac{3}{4}pi+2kpi vee x=frac{7}{4}pi+2kpi vee x=frac{5}{4}pi+2kpi\x=frac{pi}{4}+kpi vee x=frac{3}{4}pi+kpi[/latex] c) [latex]tg^2x-1=0\tg^2x=1\tgx=1 vee tgx=-1\x=frac{pi}{4}+kpi vee x=frac{3}{4}pi+kpi[/latex] d) [latex]4cos^2x-3=0\cos^2x=frac{3}{4}\cosx=frac{sqrt{3}}{2} vee cosx=-frac{sqrt{3}}{2}\x=frac{pi}{6}+2kpi vee x=frac{11}{6}pi+2kpi vee x=frac{5}{6}pi+2kpi vee x=frac{7}{6}pi+2kpi\x=frac{pi}{6}pi+kpi vee x=frac{5}{6}pi+kpi[/latex]

Dodaj swoją odpowiedź

Chemia