Задание ЕГЭ
Найдите tgα, если \frac{6sin\alpha–2cos\alpha}{4sin\alpha–4cos\alpha}=–1.Решение
Решение:
\frac{6sin\alpha–2cos\alpha}{4sin\alpha–4cos\alpha}=–1\\\frac{6sin\alpha–2cos\alpha}{4sin\alpha–4cos\alpha}=\frac{–1}{1}\\(6sin\alpha–2cos\alpha)\cdot 1=(4sin\alpha–4cos\alpha)\cdot (–1)\\6sin\alpha–2cos\alpha=–4sin\alpha+4cos\alpha\\6sin\alpha+4sin\alpha=4cos\alpha+2cos\alpha\\10sin\alpha=6cos\alpha\:{\color{Blue} |: 2}\\5sin\alpha=3cos\alpha\:{\color{Blue} |: 5cos\alpha}\\\frac{5sin\alpha}{5cos\alpha}=\frac{3cos\alpha}{5cos\alpha}\\tg\alpha=\frac{3}{5}=0,6
Ответ: 0,6.