Question

Помогите решить тригоном. ур-е : 3+9cosx=5sin^2 x

Answer 1

$5 sin^{2} x-9cosx-3=0 \\ 5(1- cos^{2} x)-9cosx-3=0 \\ 5-5 cos^{2} x-9cosx-3=0 \\ 5cos^{2}x+9cosx-2=0 \\ D= \sqrt{81 +4*2*5} =11 \\ cosx= \frac{-9+11}{10} = \frac{1}{5} \\ cosx= \frac{-9-11}{10} =-2$$-1\leq cos \alpha \leq 1$$cosx= \frac{1}{5} \\ x=arccos \frac{1}{5} +2 \pi n \\ x=-arccos \frac{1}{5} +2 \pi n$