Question

решите уравнение 2sin^2x-5+5cosx=0

Answer 1

2sin²x-5+5cosx=0;⇒2-2cos²x-5+5cosx=0;⇒
2cos²x-5cosx+3=0;⇒
cosx=y;⇒2y²-5y+3=0;y₁₂=(5⁺₋√25-4·2·3)/4=(5⁺₋1)/4;
y₁=(5+1)/4=6/4;⇒cosx≠6/4;-1<cosx<1;
y₂=4/4=1;cosx=1;x= 2kπ;k∈Z;