Question

Решить неравенство: 3*cos^2 x - 5*cos x >0 Варианты ответа: 1) (2пn; п+2пn) 2) (0.5п+ 2пn; 1,5п + 2пn) 3) (2пn-п; 2пn) 4) (-0,5п + 2пn; 0.5пn + 2пn)

Answer 1

3cos²x-5cos(x)>0

cos(x)(3cos(x)-5)>0

Составляем совокупность систем:

[{cos(x)>0, -п/2+2пk<x<п/2+2пk

[{3cos(x)-5>0, нет решений

[

[{cos(x)<0, п/2+2пk<x<3п/2+2пk

[{3cos(x)-5<0, x∈R

Отсюда остается лишь одно решение:

п/2+2пk<x<3п/2+2пk

Ответ: x∈(п/2+2пk;3п/2+2пk).

Ответ: 2.

Answer 2

To solve the inequality 3cos^2(x) - 5cos(x) > 0, we can factor it as follows:

3cos(x)(cos(x) - 5/3) > 0

So the inequality is true when either:

1. cos(x) > 0 and cos(x) > 5/3, which is impossible since -1 <= cos(x) <= 1

or

2. cos(x) < 0 and cos(x) < 5/3, which gives us:

π/2 < x < π + 2πn or 2πn < x < 2π - π/2

Therefore, the solution to the inequality is:

(π/2 + 2πn, π + 2πn) U (2πn, 2π - π/2 + 2πn)

Comparing this with the given options, we see that the answer is (1) (2πn, π + 2πn).