3 sinx + cos x/ sin x + 2 cos x = 7 /5; ⇒5*(3sin x + cos x) = 7*(sin x + 2 cos x);15 sin x + 5 cos x = 7 sin x + 14 cos x;8 sin x = 9 cos x;tg x = 9/8;1)3 sin^2 x - 2 sin x cos x + 1 = 3 sin^2 x - 2 sin x cos x + sin^2 x + cos^2 x = 4 sin^2 x - 2 sin x cos x + cos ^2 x.2) 2 cos^2 x + sin x cos x + 3 = 2 cos^2 x +sin x cos x + +3sin^2 x + 3cos^2 x = 3sin^2 x + sinx cosx + 5cos ^2 x. (4sin^2 x-2sinxcosx +cos^2 x)/(3sin^2 x+sinxcosx+5cos^ x) =(4tg^2 x - 2 tg x + 1) / (3 tg^2 x + tg x + 5) == (4*(9/8)^2 - 2*(9/8) + 1) /(3*(9/8)^2 + 9/8 + 5)== (81/16 - 9/4 + 1) / (243 /64 + 9/8 +5) = =(225/16) / (635/64) =(225/16) * (64/625) = 36/25.
f(-x) = 2tg(-5x) = -2 tg(5x) нечётная
Период функции: T = π/5
2) 2sin(x+2) = -√3
sin(x+2) = -√3/2
x + 2 = (-1)^n*arcsin(-√3/2) + πn, n∈Z
x + 2 = (-1)^(n+1)*arcsin(√3/2) + πn, n∈Z
x + 2 = (-1)^(n+1)*(π/3) + πn, n∈Z
x = (-1)^(n+1)*(π/3) - 2 + πn, n∈Z
3) 4sinx+7cosx = 0 /cosx ≠ 0
4tgx + 7 = 0
tgx = - 7/4
x = arctg(-7/4) + πk, k∈Z
x = - tg(7/4) + πk, k∈Z
4) 6tg^2x - tgx - 1 = 0
D = 1 + 4*6*1 = 25
a) tgx = (1-5)12
tgx = - 1/3
x1 = - arctg(1/3) + πn, n∈Z
б) tgx = (1+5)/12
tgx = 1/2
x2 = arctg(1/2) + πk, k∈Z
5) (cos4x - cos2x)/sinx = 0.
cos4x - cos 2x = 0; sinx ≠ 0, x1 ≠ πn, n∈Z
2*[sin(4x+2x)/2 * sin(2x-4x)/2] = 0
sin3x * sin x = 0
a) sin3x = 0
3x = πk, k∈Z
x2 = (πk)/3, k∈Z
б) sinx ≠ 0
ответ: x = (πk)/3 , k∈Z
6) Решите неравенство 1-cos2x < 0.
cos2x > 1
2x = 2πm, m∈Z
x = πm, m∈Z