A) Решить уравнение: (tg^2x-1) * sqrt (13cosx) = 0; б) Найти корни уравнения, принадлежащие отрезку [-3 п; - 3 п/2].

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.