Question

решите уравнение 8arcctg^2x +2πarcctgx-π^2

Answer 1

$8 \arcctg {}^{2} (x) + 2\pi \arcctg(x) - \pi {}^{2} = 0 \\ \arcctg(x) = t, \: t \in ( 0 ; \pi ) \\ 8 {t}^{2} + 2\pi t - {\pi}^{2} = 0 \\ D = (2\pi) {}^{2} - 4 \times 8( - {\pi}^{2} ) = 4 {\pi}^{2} + 32 {\pi}^{2} = 36 {\pi}^{2} \\ \sqrt{D} = 6\pi \\ t_{1} = \frac{ - 2\pi + 6\pi}{16} = \frac{4\pi}{16} = \frac{\pi}{4} \\ t_{2} = \frac{ - 2 \pi- 6\pi}{16} = - \frac{8\pi}{16} = - \frac{\pi}{2} \\t_{2} = - \frac{\pi}{2} \Rightarrow t_{2} \in \varnothing \\ t = \frac{\pi}{4} \\ \arcctg(x) = \frac{\pi}{4} \\ x = \ctg(\frac{\pi}{4})\\ x = 1$

Ответ: 1

Answer 2

Ответ:

Объяснение:

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