Rozwiąż równanie: [latex]4^{tg^2x} [/latex]=80- [latex]2^frac{1}{cos^2x} [/latex] w zbiorze [latex] (frac{-3 pi }{2}, frac{3 pi }{2}) - { frac{{- pi }{2}, frac{ pi }{2} [/latex]} 

Rozwiąż równanie: [latex]4^{tg^2x} [/latex]=80- [latex]2^frac{1}{cos^2x} [/latex] w zbiorze [latex] (frac{-3 pi }{2}, frac{3 pi }{2}) - { frac{{- pi }{2}, frac{ pi }{2} [/latex]} 
Odpowiedź

[latex]4^{tg^2x}=80-2^{frac1{cos^2x}} \ tg^2x=frac{sin^2x}{cos^2x}=frac{1-cos^2x}{cos^2x}=frac1{cos^2x}-frac{cos^2x}{cos^2x}=frac1{cos^2x}-1 \ 4^{frac1{cos^2x}-1}=80-2^{frac1{cos^2x}} \ 4^{frac1{cos^2x}}cdot4^{-1}=80-2^{frac1{cos^2x}} \ (2^{2})^{frac1{cos^2x}}cdotfrac14=80-2^{frac1{cos^2x}} \ (2^{frac1{cos^2x}})^2cdotfrac14=80-2^{frac1{cos^2x}} \ 2^{frac1{cos^2x}}=t [/latex] [latex]t^2cdotfrac14=80-t \ frac14t^2+t-80=0 \ Delta=1^2-4cdotfrac14cdot(-80)=1+80=81 o sqrtDelta=9 \ t_1=frac{-1-9}{2cdotfrac14}=frac{-10}{frac12}=-10cdotfrac21=-20<0 \ t_2=frac{-1+9}{2cdotfrac14}=frac8{frac12}=8cdotfrac21=16 o 2^{frac1{cos^2x}}=16 \ 2^{frac1{cos^2x}}=2^4 o frac1{cos^2x}=4 \ frac1{cos^2x}=4 o cos^2x=frac14 \ cos^2x-frac14=0 \ cos^2x-(frac12)^2=0 \ (cos x-frac12)(cos x+frac12)=0 \ cos x-frac12=0 o cos x+frac12=0 [/latex] [latex]cos x=frac12 vee cos x=-frac12 \ cos x=frac12 o x=fracpi3+2kpi vee x=-fracpi3+2kpi o xin{-frac{pi}3; fracpi3} \ cos x=-frac12 o x=frac{2pi}3+2kpi vee x=-frac{2pi}3+2kpi o xin{-frac{2pi}3; frac{2pi}3} \ x=frac{2pi}3+2kpi\ k=-1 o x=frac{2pi}3+2cdot(-1)cdotpi o x=frac{2pi}3-2pi=-frac{4pi}3 \ x=-frac{2pi}3+2kpi \ k=1 o x=-frac{2pi}3+2pi o x=frac{4pi}3 [/latex] [latex]Odp. xin{-frac{4pi}3; -frac{2pi}3; -fracpi3; fracpi3; frac{2pi}3; frac{4pi}3}[/latex]

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