Sin2x = | cosx |
Первый случай, когда cosx ≥ 0, тогда
sin2x = cosx
2sinx × cosx - cosx = 0
cosx × ( 2sinx - 1 ) = 0
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cosx = 0
x = π/2 + πn, n € Z
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2sinx - 1 = 0
sinx = 1/2
x = π/6 + 2πk, k € Z
x = 5π/ 6 + 2πm, m € Z
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Второй случай, когда соsx ≤ 0, тогда
sin2x = - cosx
2sinx × cosx + cosx = 0
cosx × ( 2sinx + 1 ) = 0
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cosx = 0
х = π/2 + πn, n € Z
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2sinx + 1 = 0
sinx = -1/2
x = - π/6 + 2πl, l € Z
x = -5π/ 6 + 2πs, s € Z
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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