Question

Помогите с 10 и 11.
Во вложении

Answer 1

$10)\; \; tgx=0,4\\\\tg(\frac{5\pi }{2}+x)=tg(2\pi +\frac{\pi}{2}+x)=tg(\frac{\pi}{2}+x)=-ctgx=-\frac{1}{tgx}=\\\\=-\frac{1}{0,4}=-\frac{10}{4}=-2,5\\\\11)\; \; 5sin^2x+13cos^2x=6\; |:cos^2x\\\\5tg^2x+13=\frac{6}{cos^2x}\; ,\; \; \; [\; \frac{1}{cos^2x}=1+tg^2x\; ]\\\\5tg^2x+13=6\cdot (1+tg^2x)\\\\5tg^2x+13=6+6tg^2x\\\\6tg^2x-5tg^2x=13-6\\\\tg^2x=7$