1) 2cos²x = 1 + sinx
2(1-sin²x) = 1 + sinx
2 - 2sin²x = 1 + sinx
2sin²x + sinx - 1 = 0
sinx обозначим t, t ∈ [-1;1]
2t² + t - 1 = 0
D = 1 + 8 = 9
t = (-1±3)/4 = 1/2 или -1
sinx = 1/2 x = ((-1)^k)*arcsin(1/2) + πk x = ((-1)^k)π/6 + πk, k ∈ Z
sinx = -1 x = -π/2 + 2πk, k ∈ Z
ответ: ((-1)^k)π/6 + πk; -π/2 + 2πk, k ∈ Z
2) cos2x + sinx = 0
1 - 2sin²x + sinx = 0
2sin²x - sinx - 1 = 0
sinx обозн. t, t ∈ [-1;1]
2t² - t - 1 = 0
t = (1±3)/4 = 1 или -1/2
sinx = 1 x = π/2 + 2πk, k ∈ Z
sinx = -1/2 x = ((-1)^k)arcsin(-1/2) + πk x = ((-1)^(k+1))π/6 + πk, k ∈ Z
ответ: ((-1)^(k+1))π/6 + πk; π/2 + 2πk, k ∈ Z
3) cos2x - cosx = 0
2cos²x - 1 - cosx = 0
2cos²x - cosx - 1 = 0
cosx обозн. t, t ∈ [-1;1]
D = (1±3)/4 = -1/2 или 1
cosx = -1/2 x = ±arccos(-1/2) + 2πk x = ±2π/3 + 2πk, k ∈ Z
cosx = 1 x = 2πk, k ∈ Z
ответ: ±2π/3 + 2πk; 2πk, k ∈ Z
4) 2cos²x = 1 - sinx
2(1 - sin²x) = 1 - sinx
2 - 2sin²x = 1 - sinx
5) Если деление на три под косинусом, тогда:
cos(2x/3) - 5cos(x/3) - 2 = 0
2cos²(x/3) - 1 - 5cos(x/3) - 2 = 0
2cos²(x/3) - 5cos(x/3) - 3 = 0
cos(x/3) обозн. t, t ∈ [-1;1]
2t² - 5t - 3 = 0
D = 25 + 24 = 49
t = (5±7)/4 = 3 или -1/2 (3 не удовл)
cos(x/3) = -1/2 x/3 = ±arccos(-1/2) + 2πk x/3 = ±2π/3 + 2πk x = ±π + 6πk, k ∈ Z
ответ: ±π + 6πk, k ∈ Z
6) sin3x = cosx
По формуле приведения
cos(π/2 - 3x) = sin3x - подставим вместо sin3x
cos(π/2 - 3x) = cosx
cos(π/2 - 3x) - cosx = 0
По формуле, сделаем из суммы произведение:
-2sin((π/2 - 3x + x)/2)sin((π/2 - 3x - x)2) = 0
sin(π/4 - x)sin(π/4 - 2x) = 0
По отдельности приравниваем к нулю:
sin(π/4 - x) = 0 π/4 - x = πk -x = -π/4 + πk x = π/4 - πk, k ∈ Z
sin(π/4 - 2x) = 0 π/4 - 2x = πk -2x = -π/4 + πk x = π/8 - πk/2, k ∈ Z
ответ: π/4 - πk; π/8 - πk/2, k ∈ Z (в ответе может быть +πk, но это значения не имеет)
7) 2cos²3x + sin3x - 1 = 0
2(1-sin²3x) + sin3x - 1 = 0
2sin²3x - sin3x - 1 = 0
sin3x обозн. t, t ∈ [-1;1]
sin3x = 1 3x = π/2 + 2πk x = π/6 + 2πk/3, k ∈ Z
sin3x = -1/2 3x = ((-1)^(k+1))π/6 + πk x = ((-1)^(k+1))π/18 + πk/3, k ∈ Z
ответ: π/6 + 2πk/3; ((-1)^(k+1))π/18 + πk/3, k ∈ Z
8) 2sin²x + cos4x = 0
По формуле понижения степени: sin²x = (1-cos2x)/2 - подставляем в уравнение:
1-cos2x + cos4x = 0
cos4x - cos2x + 1 = 0
2cos²2x - 1 - cos2x + 1 = 0
2cos²2x - cos2x = 0
cos2x(2cos2x - 1) = 0
cos2x = 0 2x = π/2 + πk x = π/4 + πk/2, k ∈ Z
cos2x = 1/2 2x = ±π/3 + 2πk x = ±π/6 + πk, k ∈ Z
ответ: π/4 + πk/2; ±π/6 + πk, k ∈ Z
1) 2cos²x = 1 + sinx
2(1-sin²x) = 1 + sinx
2 - 2sin²x = 1 + sinx
2sin²x + sinx - 1 = 0
sinx обозначим t, t ∈ [-1;1]
2t² + t - 1 = 0
D = 1 + 8 = 9
t = (-1±3)/4 = 1/2 или -1
sinx = 1/2 x = ((-1)^k)*arcsin(1/2) + πk x = ((-1)^k)π/6 + πk, k ∈ Z
sinx = -1 x = -π/2 + 2πk, k ∈ Z
ответ: ((-1)^k)π/6 + πk; -π/2 + 2πk, k ∈ Z
2) cos2x + sinx = 0
1 - 2sin²x + sinx = 0
2sin²x - sinx - 1 = 0
sinx обозн. t, t ∈ [-1;1]
2t² - t - 1 = 0
D = 1 + 8 = 9
t = (1±3)/4 = 1 или -1/2
sinx = 1 x = π/2 + 2πk, k ∈ Z
sinx = -1/2 x = ((-1)^k)arcsin(-1/2) + πk x = ((-1)^(k+1))π/6 + πk, k ∈ Z
ответ: ((-1)^(k+1))π/6 + πk; π/2 + 2πk, k ∈ Z
3) cos2x - cosx = 0
2cos²x - 1 - cosx = 0
2cos²x - cosx - 1 = 0
cosx обозн. t, t ∈ [-1;1]
2t² - t - 1 = 0
D = 1 + 8 = 9
D = (1±3)/4 = -1/2 или 1
cosx = -1/2 x = ±arccos(-1/2) + 2πk x = ±2π/3 + 2πk, k ∈ Z
cosx = 1 x = 2πk, k ∈ Z
ответ: ±2π/3 + 2πk; 2πk, k ∈ Z
4) 2cos²x = 1 - sinx
2(1 - sin²x) = 1 - sinx
2 - 2sin²x = 1 - sinx
2sin²x - sinx - 1 = 0
sinx обозн. t, t ∈ [-1;1]
2t² - t - 1 = 0
D = 1 + 8 = 9
t = (1±3)/4 = 1 или -1/2
sinx = 1 x = π/2 + 2πk, k ∈ Z
sinx = -1/2 x = ((-1)^k)arcsin(-1/2) + πk x = ((-1)^(k+1))π/6 + πk, k ∈ Z
ответ: ((-1)^(k+1))π/6 + πk; π/2 + 2πk, k ∈ Z
5) Если деление на три под косинусом, тогда:
cos(2x/3) - 5cos(x/3) - 2 = 0
2cos²(x/3) - 1 - 5cos(x/3) - 2 = 0
2cos²(x/3) - 5cos(x/3) - 3 = 0
cos(x/3) обозн. t, t ∈ [-1;1]
2t² - 5t - 3 = 0
D = 25 + 24 = 49
t = (5±7)/4 = 3 или -1/2 (3 не удовл)
cos(x/3) = -1/2 x/3 = ±arccos(-1/2) + 2πk x/3 = ±2π/3 + 2πk x = ±π + 6πk, k ∈ Z
ответ: ±π + 6πk, k ∈ Z
6) sin3x = cosx
По формуле приведения
cos(π/2 - 3x) = sin3x - подставим вместо sin3x
cos(π/2 - 3x) = cosx
cos(π/2 - 3x) - cosx = 0
По формуле, сделаем из суммы произведение:
-2sin((π/2 - 3x + x)/2)sin((π/2 - 3x - x)2) = 0
sin(π/4 - x)sin(π/4 - 2x) = 0
По отдельности приравниваем к нулю:
sin(π/4 - x) = 0 π/4 - x = πk -x = -π/4 + πk x = π/4 - πk, k ∈ Z
sin(π/4 - 2x) = 0 π/4 - 2x = πk -2x = -π/4 + πk x = π/8 - πk/2, k ∈ Z
ответ: π/4 - πk; π/8 - πk/2, k ∈ Z (в ответе может быть +πk, но это значения не имеет)
7) 2cos²3x + sin3x - 1 = 0
2(1-sin²3x) + sin3x - 1 = 0
2sin²3x - sin3x - 1 = 0
sin3x обозн. t, t ∈ [-1;1]
2t² - t - 1 = 0
D = 1 + 8 = 9
t = (1±3)/4 = 1 или -1/2
sin3x = 1 3x = π/2 + 2πk x = π/6 + 2πk/3, k ∈ Z
sin3x = -1/2 3x = ((-1)^(k+1))π/6 + πk x = ((-1)^(k+1))π/18 + πk/3, k ∈ Z
ответ: π/6 + 2πk/3; ((-1)^(k+1))π/18 + πk/3, k ∈ Z
8) 2sin²x + cos4x = 0
По формуле понижения степени: sin²x = (1-cos2x)/2 - подставляем в уравнение:
1-cos2x + cos4x = 0
cos4x - cos2x + 1 = 0
2cos²2x - 1 - cos2x + 1 = 0
2cos²2x - cos2x = 0
cos2x(2cos2x - 1) = 0
cos2x = 0 2x = π/2 + πk x = π/4 + πk/2, k ∈ Z
cos2x = 1/2 2x = ±π/3 + 2πk x = ±π/6 + πk, k ∈ Z
ответ: π/4 + πk/2; ±π/6 + πk, k ∈ Z