Question

3 ЗАДАНИЕ СРОЧНО !!!!!!!!!!!!!!!!!!!!

Answer 1

Решение
1)  sin2x = cosx
2sinx*cosx - cosx = 0
cosx*(2sinx - 1 ) = 0
1) cosx = 0 
x = π/2 + πk, k ∈ Z
2)  2sinx - 1 = 0
sinx = 1/2
x = (-1)^n* arcsin(1/2) + πn, n ∈ Z
x = (-1)^n* (π/6) + πn, n ∈ Z

2)  2cos²2x + sin2x - 1 = 0
2*(1 - sin²2x) + sin2x - 1 = 0
2 - 2sin²2x + sin2x - 1 = 0
2sin²2x - sin2x - 1 = 0
sin2x = t
2t² - t - 1 = 0
D = 1 + 4*2*1 = 9
t₁ = (1 - 3)/4
t₁ = - 1/2
t₂ = (1 + 3)/4
t₂ = 1
1)  sin2x = - 1/2
2x = (-1)^n * arcsin(-1/2) + πk, k ∈ Z
2x = (-1)^(n+1) * arcsin(1/2) + πk, k ∈ Z
2x = (-1)^(n+1) * (π/6) + πk, k ∈ Z
x = (-1)^(n+1) * (π/12) + πk/2, k ∈ Z
2)  sin2x = 1
2x = π/2 + 2πn, n ∈ Z
x = π/4 + πn, n ∈ Z