Question

Это люди, не игнорируйте тригонометрические уравнения и неравенства. sinx+1/2=0 2sin^2x-cos2x=1 ctg^2x=3 sin^2x-4sinx =5 2sin2x*cos2x-1=0 tgx/2=корень из 3 cos^2x-sin^2x=-1/2 ctg(n/2 x-n)=1

Answer 1

1)  Sinx+1/2 = 0
sinx = - 1/2
x = (-1)^n*arcsin(-1/2) + πn, n∈Z 
x = (-1)^(n + 1)*arcsin(1/2) + πn, n∈Z
 x = (-1)^(n + 1)*(π/6) + πn, n∈Z 

2)  2sin^2x - cos2x=1
 2sin^2x - (1 - 2 sin^2x)  = 1
4sin^2x - 2 = 0
sin^2x = 2/4
a)  sinx = - 1/2
x = (-1)^n*arcsin(-1/2) + πn, n∈Z
x =  (-1)^(n+1)*arcsin(1/2) + πn, n∈Z
x1 =  (-1)^(n+1)*(π/6) + πn, n∈Z
b)  sinx = 1/2
x =  (-1)^(n)*arcsin(1/2) + πk,  n∈Z
x2 =  (-1)^(n)*(π/6) + πk, k∈Z

3)  Ctg^2x=3
a)  ctgx = - √3
x1 = 5π/6 + πn, n∈Z
b)  ctgx = √3
x2 = π/6 + πk, k∈Z

4)  Sin^2x - 4sinx = 5
 Sin^2x - 4sinx - 5 = 0
sinx = t
t^2 - 4t - 5 = 0
D = 16 + 4*1*5 = 36
t1 = (4 - 6)/2
t1 = - 1
t2 = (4 + 6)/2
t2 = 5  
a)  sinx = - 1
x = - π/2 + 2πn, n∈Z
sinx = 5 не удовлетворяет условию:   I sinx I ≤ 1

5)  2sin2x*cos2x - 1= 0
sin(4x) - 1 = 0
sin(4x) = 1
4x = π/2 + 2πn, n∈Z
x = π/8 + πn/2, n∈z

6)  tg(x/2) = √3
x/2 = arctg(√3) + πn, n∈Z
x/2 = π/3 + πn, n∈Z
x = 2π/3 + 2πn, n∈Z

7)   Cos^2x-sin^2x=-1/2
cos(2x) = -1/2
2x = (+ -)*arccos(-1/2) + 2πn, n∈Z
2x = (+ -)*(π - arccos(1/2)) + 2πn, n∈Z
2x =   (+ -)*(π - π/3) + 2πn, n∈Z
2x =   (+ -)*(2π/3) + 2πn, n∈Z
x =   (+ -)*(π/3) + πn, n∈Z

8)   Ctg(n/2 x-n) = 1
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