Question

Решите срочно
1-cosx-sinx/2=0

Answer 1

sin²x/2+cos²x/2-cos²x/2+sin²x/2-sinx/2=0
2sin²x/2-sinx/2=0
sinx/2(2sinx/2 -1)=0
sinx/2=0⇒x/2=πn⇒x=2πn
sinx/2=1/2⇒x/2=(-1)^n *π/6+πn⇒x=(-1)^n*π/3+2πn

Answer 2

1)  1-cosx-sinx/2=0
(1-cosx) - sinx/2 = 0
2*sin^2(x/2) - sin(x/2) = 0
sin(x/2)*(2* sin(x/2) - 1 ) = 0
1)  sin(x/2) = 0
 x/2 = πn, n∈Z
x1 = 2πn, n∈Z

2)  2* sin(x/2) - 1 = 0
sin(x/2) = 1/2
x/2 = (-1)^n*arcsin(1/2) + πk, k∈Z
x/2 = (-1)^n*(π/6) + πk, k∈Z
x2 = (-1)^n*(π/3) + 2πk, k∈Z