Question

решите уравнение sin2x=|cos x|

Answer 1

sin2x = | cosx |

Первый случай, когда cosx ≥ 0, тогда
sin2x = cosx
2sinx × cosx - cosx = 0
cosx × ( 2sinx - 1 ) = 0
________________
cosx = 0
x = π/2 + πn, n € Z
________________
2sinx - 1 = 0
sinx = 1/2
x = π/6 + 2πk, k € Z
x = 5π/ 6 + 2πm, m € Z
_________________

Второй случай, когда соsx ≤ 0, тогда
sin2x = - cosx
2sinx × cosx + cosx = 0
cosx × ( 2sinx + 1 ) = 0
________________
cosx = 0
х = π/2 + πn, n € Z
________________
2sinx + 1 = 0
sinx = -1/2
x = - π/6 + 2πl, l € Z
x = -5π/ 6 + 2πs, s € Z
_________________

Ответ: 1) Если cosx ≥ 0, то
х = π/2 + πn, n € Z
x = π/6 + 2πk, k € Z
x = 5π/ 6 + 2πm, m € Z

2) Если cosx ≤ 0 , то
х = π/2 + πn, n € Z
x = - π/6 + 2πl, l € Z
x = -5π/ 6 + 2πs, s € Z