Question

Помогите с алгеброй

Answer 1

Ответ:

a) (tg^2x - 1) * sqrt(13cosx) = 0;

tg^2x - 1 = 0 или sqrt(13cosx) = 0;

tg^2x = 1

tgx = +-1

x = pi/4 + pin, n ∈ Z;

x = -pi/4 + pin, n ∈ Z;

sqrt(13cosx) = 0;

13cosx = 0;

cosx = 0;

x = pi/2 + pin, n ∈ Z;

б) -3pi <= pi/4 + pin <= -3pi/2;

-12pi/4 - pi/4 <= pin <= -6pi/4 - pi/4;

-13pi/4 <= pin <= -7pi/4;

n = -3, n = -2: x = pi/2 - 2pi = -3pi/2; x = pi/2 - 3pi = -5pi/2;

-3pi <= pi/2 + pin <= -3pi/2;

-3pi - pi/2 <= pin <= -3pi/2 - pi/2;

-7pi/2 <= pin <= -4pi/2;

-3,5 <= n <= -2;

n = -3, n = -2: x = pi/2 - 3pi = -5pi/2, x = pi/2 - 2pi = -3pi/2.