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Решите задачу:

$8 \sin(x) - 1 = 4(1 - { \sin(x) }^{2}) \\ \sin(x) = t \\ 8t - 1 = 4 - 4 {t}^{2} \\ 4 {t}^{2} + 8t - 5 = 0 \\ d = 64 - 4 \times 4 \times ( - 5) = 64 + 80 = 144 \\ t1 = \frac{ - 8 - 12}{8} = - \frac{5}{2} \\ t2 = \frac{ - 8+ 12}{8} = \frac{4}{8} = \frac{1}{2} \\ \sin(x) = - \frac{5}{2} \: \: eto \: newozmozhna \: \\ \sin(x) = \frac{1}{2} \\ x = \frac{\pi}{6} + \pi \times k$