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Решите задачу:

$1)\; \; \; 5sin^2 \alpha +13cos^2 \alpha =6\; |:cos^2 \alpha \ne 0\\\\5tg^2 \alpha +13=6\cdot \frac{1}{cos^2 \alpha }\\\\5tg^2 \alpha +13=6\cdot (1+tg^2 \alpha )\\\\5tg^2 \alpha +13=6+6tg^2 \alpha \\\\tg^2 \alpha =7\\\\2)\; \; tg \alpha =2\\\\\frac{2cos \alpha -7sin \alpha }{2sin \alpha -2cos \alpha } = \frac{cos \alpha (2-7tg \alpha )}{cos \alpha (2tg \alpha -2)} = \frac{2-7\cdot 2}{2\cdot 2-2} = \frac{-12}{2} =-6$

$3)\; \; cos \beta =-\frac{3}{5}\\\\3cos(\pi + \beta )+2sin(\frac{3\pi}{2}+ \beta )=-3cos-2cos \beta =-5cos \beta =\\\\=-5\cdot (-\frac{3}{5})=3$