Rozwiąż równanie:=80- w zbiorze }.... Question from @Madzia765

December 2018 1 164 Report

Rozwiąż równanie:
$4^{tg^2x}$ =80- $2^\frac{1}{cos^2x}$ w zbiorze $(\frac{-3 \pi }{2}, \frac{3 \pi }{2}) - { \frac{{- \pi }{2}, \frac{ \pi }{2}$ }

loitzl9006 $4^{tg^2x}=80-2^{\frac1{\cos^2x}} \\ tg^2x=\frac{\sin^2x}{\cos^2x}=\frac{1-\cos^2x}{\cos^2x}=\frac1{\cos^2x}-\frac{\cos^2x}{\cos^2x}=\frac1{\cos^2x}-1 \\ 4^{\frac1{\cos^2x}-1}=80-2^{\frac1{\cos^2x}} \\ 4^{\frac1{\cos^2x}}\cdot4^{-1}=80-2^{\frac1{\cos^2x}} \\ (2^{2})^{\frac1{\cos^2x}}\cdot\frac14=80-2^{\frac1{\cos^2x}} \\ (2^{\frac1{\cos^2x}})^2\cdot\frac14=80-2^{\frac1{\cos^2x}} \\ 2^{\frac1{\cos^2x}}=t$
$t^2\cdot\frac14=80-t \\ \frac14t^2+t-80=0 \\ \Delta=1^2-4\cdot\frac14\cdot(-80)=1+80=81 \ \to \ \sqrt\Delta=9 \\ t_1=\frac{-1-9}{2\cdot\frac14}=\frac{-10}{\frac12}=-10\cdot\frac21=-20<0 \\ t_2=\frac{-1+9}{2\cdot\frac14}=\frac8{\frac12}=8\cdot\frac21=16 \ \to \ 2^{\frac1{\cos^2x}}=16 \\ 2^{\frac1{\cos^2x}}=2^4 \ \to \ \frac1{\cos^2x}=4 \\ \frac1{\cos^2x}=4 \ \to \ \cos^2x=\frac14 \\ \cos^2x-\frac14=0 \\ \cos^2x-(\frac12)^2=0 \\ (\cos x-\frac12)(\cos x+\frac12)=0 \\ \cos x-\frac12=0 \ \to \ \cos x+\frac12=0$
$\cos x=\frac12 \ \vee \ \cos x=-\frac12 \\ \cos x=\frac12 \ \to \ x=\frac\pi3+2k\pi \ \vee \ x=-\frac\pi3+2k\pi \ \to \ x\in\{-\frac{\pi}3; \ \frac\pi3\} \\ \cos x=-\frac12 \ \to \ x=\frac{2\pi}3+2k\pi \ \vee \ x=-\frac{2\pi}3+2k\pi \ \to \ x\in\{-\frac{2\pi}3; \ \frac{2\pi}3\} \\ x=\frac{2\pi}3+2k\pi\\ k=-1 \ \to x=\frac{2\pi}3+2\cdot(-1)\cdot\pi \ \to \ x=\frac{2\pi}3-2\pi=-\frac{4\pi}3 \\ x=-\frac{2\pi}3+2k\pi \\ k=1 \ \to \ x=-\frac{2\pi}3+2\pi \ \to x=\frac{4\pi}3$
$Odp. \ x\in\{-\frac{4\pi}3; \ -\frac{2\pi}3; \ -\frac\pi3; \ \frac\pi3; \ \frac{2\pi}3; \ \frac{4\pi}3\}$

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