Odpowiedź:
[tex]\huge\boxed{a) \ \alpha = 60^{o}}\\\\\\\huge\boxed{b) \ \alpha = 30^{o}}\\\\\\\huge\boxed{c) \ \alpha = 45^{o}}\\\\\\\huge\boxed{d) \ \alpha = 30^{o}}}[/tex]
Szczegółowe wyjaśnienie:
Korzystamy z zależności trygonometrycznych:
[tex]sin^{2}\alpha + cos^{2}\alpha = 1 \ \ \rightarrow \ \ sin^{2}\alpha=1-cos^{2}\alpha \ \ \rightarrow \ \ cos^{2}\alpha =1-sin^{2}\alpha\\\\tg\alpha \cdot ctg\alpha = 1[/tex]
Pamiętamy, że dla kąta ostrego wszystkie wartości funkcji są dodatnie.
[tex]a)\\6sin^{2}\alpha = 5 - cos\alpha\\\\6(1-cos^{2}\alpha)=5-cos\alpha\\\\6-6cos^{2}\alpha -5+cos\alpha = 0\\\\-6cos^{2}\alpha + cos\alpha +1 = 0 \ \ \ |\cdot(-1)\\\\6cos^{2}\alpha - cos\alpha -1 = 0\\\\a = 6, \ b = -1, \ c = -1\\\\\Delta = b^{2}-4ac = (-1)^{2}-4\cdot6\cdot(-1) = 1+24 = 25\\\\\sqrt{\Delta} = \sqrt{25}} = 5\\\\cos\alpha=\frac{-b-\sqrt{\Delta}}{2a} = \frac{1-5}{2\cdot6}=\frac{-4}{12} = -\frac{1}{3} \ \ < 0, \ odpada[/tex]
[tex]lub\\\\cos\alpha = \frac{-b+\sqrt{\Delta}}{2a} = \frac{1+5}{12} = \frac{6}{12} = \frac{1}{2} \ \ \rightarrow \ \ \boxed{\alpha = 60^{o}}[/tex]
[tex]b)\\2cos^{2}\alpha = 5sin\alpha - 1\\\\2(1-sin^{2}\alpha)=5sin\alpha -1\\\\2-2sin^{2}\alpha - 5sin\alpha+1 = 0\\\\-2sin^{2}\alpha -5sin\alpha+3 = 0 \ \ \ |\cdot(-1)\\\\2sin^{2}\alpha + 5sin\alpha - 3 = 0\\\\\Delta = 5^{2}-4\cdot2\cdot(-3) = 25 + 24 = 49\\\\\sqrt{\Delta} = \sqrt{49} = 7\\\\sin\alpha = \frac{-5-7}{2\cdot2} = \frac{-12}{4} = -3, \ \ odpada[/tex]
[tex]lub\\\\sin\alpha = \frac{-5+7}{4} = \frac{2}{4} = \frac{1}{2} \ \ \rightarrow \ \ \boxed{ \alpha = 30^{o}}[/tex]
[tex]c)\\tg\alpha + ctg\alpha = 2\\\\tg\alpha + \frac{1}{tg\alpha}-2 = 0 \ \ \ |\cdot tg\alpha\\\\tg^{2}\alpha + 1 - 2tg\alpha = 0\\\\tg^{2}\alpha -2tg\alpha + 1 = 0\\\\(tg\alpha -1)^{2} = 0\\\\tg\alpha -1 = 0\\\\tg\alpha = 1 \ \ \rightarrow \ \ \boxed{\alpha = 45^{o}}[/tex]
[tex]d)\\3tg^{2}\alpha + 2\sqrt{3}tg\alpha = 3\\\\3tg^{2}\alpha + 2\sqrt{3}tg\alpha -3 = 0\\\\\Delta = (2\sqrt{3})^{2}-4\cdot3\cdot(-3) = 12+36 = 48\\\\\sqrt{\Delta} = \sqrt{48} = \sqrt{16\cdot3} = 4\sqrt{3}[/tex]
[tex]tg\alpha = \frac{-2\sqrt{3}-4\sqrt{3}}{2\cdot3} = \frac{-6\sqrt{3}}{6} = -\sqrt{3} \ \ < 0, \ \ odpada\\\\lub\\\\tg\alpha = \frac{-2\sqrt{3}+4\sqrt{3}}{6} = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3} \ \ \rightarrow \ \ \boxed{\alpha = 30^{o}}[/tex]
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Verified answer
Odpowiedź:
[tex]\huge\boxed{a) \ \alpha = 60^{o}}\\\\\\\huge\boxed{b) \ \alpha = 30^{o}}\\\\\\\huge\boxed{c) \ \alpha = 45^{o}}\\\\\\\huge\boxed{d) \ \alpha = 30^{o}}}[/tex]
Szczegółowe wyjaśnienie:
Korzystamy z zależności trygonometrycznych:
[tex]sin^{2}\alpha + cos^{2}\alpha = 1 \ \ \rightarrow \ \ sin^{2}\alpha=1-cos^{2}\alpha \ \ \rightarrow \ \ cos^{2}\alpha =1-sin^{2}\alpha\\\\tg\alpha \cdot ctg\alpha = 1[/tex]
Pamiętamy, że dla kąta ostrego wszystkie wartości funkcji są dodatnie.
[tex]a)\\6sin^{2}\alpha = 5 - cos\alpha\\\\6(1-cos^{2}\alpha)=5-cos\alpha\\\\6-6cos^{2}\alpha -5+cos\alpha = 0\\\\-6cos^{2}\alpha + cos\alpha +1 = 0 \ \ \ |\cdot(-1)\\\\6cos^{2}\alpha - cos\alpha -1 = 0\\\\a = 6, \ b = -1, \ c = -1\\\\\Delta = b^{2}-4ac = (-1)^{2}-4\cdot6\cdot(-1) = 1+24 = 25\\\\\sqrt{\Delta} = \sqrt{25}} = 5\\\\cos\alpha=\frac{-b-\sqrt{\Delta}}{2a} = \frac{1-5}{2\cdot6}=\frac{-4}{12} = -\frac{1}{3} \ \ < 0, \ odpada[/tex]
[tex]lub\\\\cos\alpha = \frac{-b+\sqrt{\Delta}}{2a} = \frac{1+5}{12} = \frac{6}{12} = \frac{1}{2} \ \ \rightarrow \ \ \boxed{\alpha = 60^{o}}[/tex]
[tex]b)\\2cos^{2}\alpha = 5sin\alpha - 1\\\\2(1-sin^{2}\alpha)=5sin\alpha -1\\\\2-2sin^{2}\alpha - 5sin\alpha+1 = 0\\\\-2sin^{2}\alpha -5sin\alpha+3 = 0 \ \ \ |\cdot(-1)\\\\2sin^{2}\alpha + 5sin\alpha - 3 = 0\\\\\Delta = 5^{2}-4\cdot2\cdot(-3) = 25 + 24 = 49\\\\\sqrt{\Delta} = \sqrt{49} = 7\\\\sin\alpha = \frac{-5-7}{2\cdot2} = \frac{-12}{4} = -3, \ \ odpada[/tex]
[tex]lub\\\\sin\alpha = \frac{-5+7}{4} = \frac{2}{4} = \frac{1}{2} \ \ \rightarrow \ \ \boxed{ \alpha = 30^{o}}[/tex]
[tex]c)\\tg\alpha + ctg\alpha = 2\\\\tg\alpha + \frac{1}{tg\alpha}-2 = 0 \ \ \ |\cdot tg\alpha\\\\tg^{2}\alpha + 1 - 2tg\alpha = 0\\\\tg^{2}\alpha -2tg\alpha + 1 = 0\\\\(tg\alpha -1)^{2} = 0\\\\tg\alpha -1 = 0\\\\tg\alpha = 1 \ \ \rightarrow \ \ \boxed{\alpha = 45^{o}}[/tex]
[tex]d)\\3tg^{2}\alpha + 2\sqrt{3}tg\alpha = 3\\\\3tg^{2}\alpha + 2\sqrt{3}tg\alpha -3 = 0\\\\\Delta = (2\sqrt{3})^{2}-4\cdot3\cdot(-3) = 12+36 = 48\\\\\sqrt{\Delta} = \sqrt{48} = \sqrt{16\cdot3} = 4\sqrt{3}[/tex]
[tex]tg\alpha = \frac{-2\sqrt{3}-4\sqrt{3}}{2\cdot3} = \frac{-6\sqrt{3}}{6} = -\sqrt{3} \ \ < 0, \ \ odpada\\\\lub\\\\tg\alpha = \frac{-2\sqrt{3}+4\sqrt{3}}{6} = \frac{2\sqrt{3}}{6} = \frac{\sqrt{3}}{3} \ \ \rightarrow \ \ \boxed{\alpha = 30^{o}}[/tex]