помогите!срочно! найдите все общие корни уравнений 5cos2x+2cosx-3=0 и sin2x+14cos^2x-8=0

Ответы 1

$1) \\ 5 \cos(2x) + 2 \cos(x) - 3 = 0 \\ 5(2 \cos {}^{2} (x) - 1) + 2 \cos(x) - 3 = 0 \\ 10 { \cos {}^{2} ( x) } - 5 + 2 \cos(x) - 3 = 0 \\ 10 \cos {}^{2} (x) + 2 \cos(x) - 8 = 0 \\ 5 \cos {}^{2} (x) + \cos(x ) - 4 = 0 \\ d = 1 + 4 \times 4 \times 5 = 81 = {9}^{2} \\ \cos(x) = \frac{ -1 ±9}{10} = - 1 ;\frac{4}{5} \\ x = \pi + 2\pi \: k \: \\ x = ±arccos( \frac{4}{5} ) + 2\pi \: k \\$ $\sin(2x) + 14 \cos {}^{2} (x) - 8 = 0 \\ 14 \cos {}^{2} (x) + 2 \sin(x) \cos(x) - 8 \cos {}^{2} (x) - \sin {}^{2} (x) = 0 \\ 6 \cos {}^{2} (x) + 2 \sin(x) \cos(x) - 8 \sin {}^{2} (x) = 0 \: \: / \div \cos {}^{2} (x) \\ 6 + 2 \tan(x) - 8 \tan {}^{2} (x) = 0 \\ 8 \tan {}^{2} (x) - 2 \tan(x) - 6 = 0/ \div 2 \\ 4 \tan {}^{2} (x) - \tan(x) - 3 = 0 \\ d = 1 + 4 \times 3 \times 4 = 49 = {7}^{2} \\ \tan(x) = \frac{1±7}{8} = 1; - \frac{3}{4} \\ x = \frac{\pi}{4} + \pi \: k \\ x = - arctg( \frac{3}{4} ) + \pi \: k$
- Автор:
  ada
- 5 лет назад
- 0
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