Question

4sin ^2 x +4 cos x -1=0

Answer 1

Решение
4sin∧2x + 4cosx - 1 = 0
4*(1 - cos∧2x)  + 4 cosx - 1 = 0
4 - 4cos∧2x + 4cosx - 1 = 0
4cos∧2x - 4cosx - 3 = 0
D = 16 + 4*4*3 = 64
1) cosx = (4 - 8) / 8 = -1/2
 cosx = -1/2
x = (+ -) arccos(-1/2) + 2πn, n∈Z
x = (+ -) (π - π/3) + 2πn, n∈Z
x =( + -)(2π/3) + 2πn, n∈Z
2) cosx = (4 + 8 ) / 8 = 3/2 не удовлетворяет области определения функции y = cosx (  -1 ≤ cosx ≤ 1)