Question

Решите a) 2 cos^2x - 3 cos x + 1 = 0; б) 2 cos^2x - 2 cos x - 1 =0; в) 2 sin^2x + sin x - 1 = 0; г) 6 tg^2x + tg x - 1 = 0. а) 3 sin^2 x + sinx × cosx = 2 cos^2 x; б) 2 cos^2 - 3 sinx × cosx + sin^2x = 0; в) 9 sin x cos x - 7cos^2x = 2 sin^2x; г) 2 sin^2 x - sinx cosx = cos^2 x.

Answer 1

2cos^2 x - 3cos x + 1 = 0;

cos x=a, -1≤a≤1,

2a^2-3a+1=0,

D=1,

a1=1/2, a2=1,

cos x=1/2,

x=±arccos(1/2)+2πk, k∈Z,

x=±π/3+2πk, k∈Z,

cos x=1,

x=2πk, k∈Z;

3sin^2 x + sinx × cosx = 2cos^2 x;

3sin^2 x + sinx × cosx - 2cos^2 x=0;

3(sin x/cos x)^2 + sin x/cos x -2=0,

3tg^2 x + tg x - 2=0,

tg x=a,

3a^2+a-2=0,

D=25,

a1=-1, a2=2/3,

tg x=-1,

x=arctg(-1)+πk, k∈Z,

x=-arctg1+πk, k∈Z,

x=-π/4+πk, k∈Z,

x=arctg(2/3)+πk, k∈Z.