Odpowiedź:
Zad.4
a)
[tex]-x^2-3x+2\leq 0\\x^2+3x-2\geq 0\\ \Delta=9+8=17\\\sqrt{\Delta}=\sqrt{17}\\x_1=\frac{-3-\sqrt{17}}{2}\\x_2=\frac{-3+\sqrt{17}}{2}=\frac{\sqrt{17}-3}{2}\\\\x\in(-\infty,\frac{-3-\sqrt{17}}{2}\rangle\cup\langle\frac{\sqrt{17}-3}{2},+\infty)[/tex]
b)
[tex]2x^2+5x+2 < 0\\\Delta=25-16=9\\\sqrt{\Delta}=3\\x_1=\frac{-5-3}{4}=-2\\x_2=\frac{-5+3}{4}=-\frac{1}{2}\\\\x\in(-2,-\frac{1}{2})[/tex]
Zad.5
[tex]sin\alpha=\frac{3}{5}, \ \alpha\in(0,\frac{\pi}{2})\\\\sin^2\alpha+cos^2\alpha=1\\(\frac{3}{5})^2+cos^2\alpha=1\\\frac{9}{25}+cos^2\alpha=1\\cos^2\alpha=\frac{16}{25}\\cos\alpha=\frac{4}{5}, \ \alpha\in(0,\frac{\pi}{2})\\\\tg\alpha=\frac{sin\alpha}{cos\alpha}\\tg\alpha=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{5}*\frac{5}{4}=\frac{3}{4}\\\\ctg\alpha=\frac{1}{tg\alpha}\\ctg\alpha=\frac{1}{\frac{3}{4}}=\frac{4}{3}[/tex]
[tex]tg\alpha=2, \ \alpha\in (0,\frac{\pi}{2})\\\\tg\alpha=\frac{sin\alpha}{cos\alpha} \implies sin\alpha=tg\alpha cos\alpha\\sin\alpha =2cos\alpha \\\\sin^2\alpha +cos^2\alpha =1\\(2cos\alpha)^2+cos^2\alpha =1\\4cos^2\alpha +cos^2\alpha=1\\5cos^2\alpha =1\\cos^2\alpha =\frac{1}{5}\\cos\alpha=\frac{\sqrt{5}}{5}, \ \alpha\in (0,\frac{\pi}{2})\\\\sin\alpha =2cos\alpha \\sin\alpha =2*\frac{\sqrt{5}}{5}=\frac{2\sqrt{5}}{5}\\\\ctg\alpha=\frac{1}{tg\alpha}\\ctg\alpha =\frac{1}{2}[/tex]
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Odpowiedź:
Zad.4
a)
[tex]-x^2-3x+2\leq 0\\x^2+3x-2\geq 0\\ \Delta=9+8=17\\\sqrt{\Delta}=\sqrt{17}\\x_1=\frac{-3-\sqrt{17}}{2}\\x_2=\frac{-3+\sqrt{17}}{2}=\frac{\sqrt{17}-3}{2}\\\\x\in(-\infty,\frac{-3-\sqrt{17}}{2}\rangle\cup\langle\frac{\sqrt{17}-3}{2},+\infty)[/tex]
b)
[tex]2x^2+5x+2 < 0\\\Delta=25-16=9\\\sqrt{\Delta}=3\\x_1=\frac{-5-3}{4}=-2\\x_2=\frac{-5+3}{4}=-\frac{1}{2}\\\\x\in(-2,-\frac{1}{2})[/tex]
Zad.5
a)
[tex]sin\alpha=\frac{3}{5}, \ \alpha\in(0,\frac{\pi}{2})\\\\sin^2\alpha+cos^2\alpha=1\\(\frac{3}{5})^2+cos^2\alpha=1\\\frac{9}{25}+cos^2\alpha=1\\cos^2\alpha=\frac{16}{25}\\cos\alpha=\frac{4}{5}, \ \alpha\in(0,\frac{\pi}{2})\\\\tg\alpha=\frac{sin\alpha}{cos\alpha}\\tg\alpha=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{5}*\frac{5}{4}=\frac{3}{4}\\\\ctg\alpha=\frac{1}{tg\alpha}\\ctg\alpha=\frac{1}{\frac{3}{4}}=\frac{4}{3}[/tex]
b)
[tex]tg\alpha=2, \ \alpha\in (0,\frac{\pi}{2})\\\\tg\alpha=\frac{sin\alpha}{cos\alpha} \implies sin\alpha=tg\alpha cos\alpha\\sin\alpha =2cos\alpha \\\\sin^2\alpha +cos^2\alpha =1\\(2cos\alpha)^2+cos^2\alpha =1\\4cos^2\alpha +cos^2\alpha=1\\5cos^2\alpha =1\\cos^2\alpha =\frac{1}{5}\\cos\alpha=\frac{\sqrt{5}}{5}, \ \alpha\in (0,\frac{\pi}{2})\\\\sin\alpha =2cos\alpha \\sin\alpha =2*\frac{\sqrt{5}}{5}=\frac{2\sqrt{5}}{5}\\\\ctg\alpha=\frac{1}{tg\alpha}\\ctg\alpha =\frac{1}{2}[/tex]