Найдите $\tg^2\alpha $, если 5sin2α+13cos2α=65\sin^2\alpha +13\cos^2\alpha =65sin2α+13cos2α=6. Источник: Банк ФИПИ Показать решение Решение: 5sin2α+13cos2α=6 ⟹ 5(1−cos2α)+13cos2α=65\sin^2\alpha + 13\cos^2\alpha = 6 \implies 5(1-\cos^2\alpha) + 13\cos^2\alpha = 65sin2α+13cos2α=6⟹5(1−cos2α)+13cos2α=6 5−5cos2α+13cos2α=6 ⟹ 8cos2α=1 ⟹ cos2α=185 - 5\cos^2\alpha + 13\cos^2\alpha = 6 \implies 8\cos^2\alpha = 1 \implies \cos^2\alpha = \frac{1}{8}5−5cos2α+13cos2α=6⟹8cos2α=1⟹cos2α=81 sin2α=78 ⟹ tg2α=sin2αcos2α=7\sin^2\alpha = \frac{7}{8} \implies \tg^2\alpha = \frac{\sin^2\alpha}{\cos^2\alpha} = 7sin2α=87⟹tg2α=cos2αsin2α=7 Ответ: 7