Найдите все решения уравнения на промежутке [[tex]-2\pi[/tex];[tex]2\pi[/tex]]3cos5x-3[tex]\sqrt{2}[/tex] = 3sin5x +[tex]\sqrt{2}[/tex] sin^2(x+[tex]\frac{\pi }{20}[/tex])

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Для начала преобразуем уравнение:

3cos(5x) - 3√2 = 3sin(5x) + √2sin^2(x + π/20)

3cos(5x) - 3√2 = 3sin(5x) + √2(sin(x)cos(π/20) + cos(x)sin(π/20))

3cos(5x) - 3√2 = 3sin(5x) + √2(sin(x)cos(π/20) + (cos(x)sin(π/20))

3cos(5x) - 3√2 = 3sin(5x) + √2sin(x)cos(π/20) + √2cos(x)sin(π/20)

Переносим все слагаемые в левую часть:

3cos(5x) - 3sin(5x) - √2sin(x)cos(π/20) - √2cos(x)sin(π/20) - 3√2 = 0

Раскроем тригонометрические функции:

3cos(5x) - 3sin(5x) - √2(1/2sin(x)) - √2(1/2cos(x)) - 3√2 = 0

3cos(5x) - 3sin(5x) - √2/2sin(x) - √2/2cos(x) - 3√2 = 0

3cos(5x) - 3sin(5x) - √2/2 * (sin(x) + cos(x)) - 3√2 = 0

3cos(5x) - 3sin(5x) - √2/2 √2 sin(x + π/4) - 3√2 = 0

3cos(5x) - 3sin(5x) - sin(x + π/4) - 3√2 = 0

3cos(5x) - 3sin(5x) - sin(x)cos(π/4) - cos(x)sin(π/4) - 3√2 = 0

3cos(5x) - 3sin(5x) - √2/2(sin(x) + cos(x)) - 3√2 = 0

3cos(5x) - 3sin(5x) - √2/2√2 * (sin(x) + cos(x)) - 3√2 = 0

3cos(5x) - 3sin(5x) - sin(x + π/4) - 3√2 = 0

3(cos(5x) - sin(5x)) = (1 + 3√2)sin(x + π/4)

Выразим sin(5x) и cos(5x) через sin(x) и cos(x):

cos(5x) = cos^2(2x) - sin^2(2x) = (1 - 2sin^2(x))(1 - 2cos^2(x)) - 2cos(x)sin(x)(2cos(x)sin(x)) = 1 - 2(sin^2(x) + cos^2(x) - 2cos(x)sin(x)) = 1 - 2(1 - 2sin(x)cos(x) - 2cos(x)sin(x)) = 1 + 4sin(x)cos(x) - 4cos(x)sin(x) - 2 = 2 - 4sin(x)cos(x)

sin(5x) = sin(x)cos(4x) + cos(x)sin(4x) = sin(x)(cos^4(x) - 6sin^2(x)cos^2(x) + sin^4(x)) + cos(x)(1 - 8sin^2(x)cos^2(x)) = sin(x)(1 - 2sin^2(x))(1 - 2cos^2(x)) + cos(x)(1 - 8sin^2(x)cos^2(x)) = sin(x)(1 - 2 + 4sin^2(x) - 4sin^2(x) - 2cos^2(x)) + cos(x)(1 - 8sin^2(x) + 8sin^2(x)cos(x)^2) = sin(x)(-1 - 2cos^2(x)) + cos(x)(1 - 8sin^2(x) + 8sin^2(x)(1 - sin^2(x))) = -sin(x)(1 + 2 - 2sin^2(x)) + cos(x)(1 - 8sin^2(x) + 8sin^2(x) - 8sin^4(x)) = -sin(x)(1 + 2cos^2(x)) + cos(x)(1 + 16sin^2(x) - 8sin^4(x))

Подсчитаем:

2cos(5x) = 2(2 - 4sin(x)cos(x)) = 4 - 8sin(x)cos(x)

sin(5x) = cos(x)[1 + 16sin^2(x) - 8sin^4(x)] - sin(x)[1 + 2cos^2(x)] = cos(x) - 2cos^2(x)sin(x) + 16sin^3(x)cos(x) - 8sin^5(x) - sin(x) - 16sin^3(x) + 8sin^5(x)

Теперь подставим выражения для sin(5x) и cos(5x) в уравнение:

3(4 - 8sin(x)cos(x)) - 3(cos(x) + 16sin^3(x)cos(x) - 8sin^5(x)) = (1 + 3√2)(-sin(x)cos(x) + cos(x) - 2cos^2(x)sin(x) + 16sin^3(x)cos(x) - 8sin^5(x) - sin(x) - 16sin^3(x) + 8sin^5(x))

12 - 24sin(x)cos(x) - 3cos(x) - 48sin^3(x)cos(x) + 24sin^5(x) = -sin(x)cos(x) - 3sin(x) - 6√2sin(x)cos(x) + 16√2sin^3(x)cos(x) - 8√2sin^5(x) - 2cos^2(x)sin^2(x) + 3cos(x) - 8sin^3(x)cos(x) + 32sin^5(x) + 3√2sin(x) + 48√2sin^3(x) - 24√2sin^5(x)

12cos(π/4) + 24sin(π/4) - 6√2sin(x)cos(x) + 16√2sin^3(x)cos(x) - 8√2sin^5(x) = -sin(x)cos(x) - 6√2-2sin^2(x)cos(x) + 3√2sin(x) - 8√2sin^3(x)cos(x) + 32√2sin^5(x) + 3√2sin(x) + 48√2sin^3(x) - 24√2sin^5(x)

Подставим значения sin(π/4) = cos(π/4) = √2/2:

12√2/2 + 24√2/2 - 6√2sin(x)cos(x) + 16√2sin^3(x)cos(x) - 8√2sin^5(x) = -sin(x)cos(x) - 6√2-2sin^2(x)cos(x) + 3√2sin(x) - 8√2sin^3(x)cos(x) + 32√2sin^5(x) + 3√2sin(x) + 48√2sin^3(x) - 24√2sin^5(x)
6√2 + 12√2 - 6√2sin(x)cos(x) + 16√2sin^3(x)cos(x) - 8√2sin^5(x) = -sin(x)cos(x) - 3√2sin(x)cos(x) + 3√2sin(x) - 12√2sin^3(x)cos(x) + 48√2sin^5(x) + 3√2sin(x) + 48√2sin^3(x) - 24√2sin^5(x)

18√2 - 6√2sin(x)cos(x) + 16√2sin^3(x)cos(x) - 8√2sin^5(x) = sin(x)(-cos(x) - 3cos(x) + 3 - 12cos(x) + 48sin^2(x) + 3 + 48sin^2(x) - 24sin^4(x))

18√2 - 6√2sin(x)cos(x) + 16√2sin^3(x)cos(x) - 8√2sin^5(x) = 18√2sin(x) - 18√2sin^3(x)

Теперь решим получившееся уравнение.