1)23cosx+19sin^2x+19cos^2x=18 2)cos^2 *2x/23=3/4 3)17sin^2x+289sinx=0

Question

1)23cosx+19sin^2x+19cos^2x=18
2)cos^2 *2x/23=3/4
3)17sin^2x+289sinx=0

Answer 1

1)23cosx+19sin²x+19cos²x=18;⇒23cosx+19(sin²x+cos²x)=18;⇒
23cosx=18-19=-1;⇒
cosx=-1/23;⇒x=⁺₋arccos(-1/23)+2kπ;k∈Z;
3)17sin²x+289sinx=0;⇒17sinx(sinx+17)=0;⇒sinx(sinx+17)=0;
sinx=0;⇒x=kπ;k∈Z;
sinx+17=0;⇒sinx=-17;⇒-1