Question

С подробным объяснением, пожалуйста
cos(x/2)=0

4+3cos2x=1

cos(6+3x)=-($\frac{ \sqrt{2}}{2}$)

Answer 1

1)  cos(x/2) = 0
x/2 = π/2 + πk, k∈Z
x = π + 2πk, k∈z

2)  4 + 3cos2x = 1
3cos2x = - 3
cos2x = - 1
2x = π + 2πn, n∈Z
x = π/2 + πn, n∈z

3)  cos(6+3x) = - ()
6 + 3x = (+ -)arccos(- √2/2) + 2πk, k∈Z
6 + 3x = (+ -)(π - arccos√2/2) + 2πk, k∈z
6 + 3x = (+ -)*(π - π/4) + 2πk, k∈Z
6 + 3x = (+ -)(3π/4) + 2πk, k∈z
3x = (+ -)*(3π/4) - 6 + 2πk, k∈z
 x = (+ -)*(π/4) - 2 + 2πk/3, k∈Z