Найдите корни уравнения заданном промежутке

2 sin ² x = 1, x принадлежит [ 0; 2π]

2 sin²x=1 x∈[0; 2π]

sin²x=1/2

sinx=√ (1/2) sinx=-√ (1/2)

sinx=1/√2 sinx=-1√2

sinx=√2 sinx=-√2

2 2

x = (-1) ^n*arcsin (√2) + πn x = (-1) ^ (n+1) * arcsin (√2) + πn

2 2

x = (-1) ^n * π+πn x = (-1) ^ (n+1) * π+πn

4 4

n=0 x = (-1) ⁰ * π = π ∈[0; 2π] x = (-1) ¹ * π = - π ∉[0; 2π]

4 4 4 4

n=1 x = (-1) ¹ * π+π=-π+π=3π∈[0; 2π] x = (-1) ² * π+π=π + π = 5π ∈[0; 2π]

4 4 4 4 4 4

n=2 x = (-1) ² * π + 2π=9π ∉[0; 2π] x = (-1) ³ π+2π=7π ∈[0; 2π]

4 4 4 4

Ответ: π; 3π; 5π; 7π

4 4 4 4

1-2sin²x=0

cos2x=0

2x=π/2+πn

x=π/4+πn/2

0≤π/4+πn/2≤2π

-π/4≤πn/2≤-π/4+2π

-1/2≤n≤3 1/2

n=0 x=π/4

n=1 x=3π/4

n=2 x=5π/4

n=3 x=7π/4