Ответ:

Объяснение:

sin x = √3/2

x = 2/3pi + 2pi*n (n - целое) и x = 1/3pi + 2pi*n (n - целое)

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cos x = 1/2

x = -1/3pi + 2pi*n (n - целое) и x = 1/3pi + 2pi*n (n - целое)

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cos x = -√3/2

x = 5/6pi + 2pi*n и x = -5/6pi + 2pi*n (n - целое)

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cos x = -1

x = pi + 2pi*n (n - целое)

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tg x = √3

x = 1/3pi + pi*n (n - целое)

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sin x = -1/2

x = -1/6pi + 2pi*n и x = 7/6pi + 2pi*n (n - целое)

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3sin^2x - 5sinx - 2 = 0

(3sinx +1)(sinx-2) = 0

sin(x) - 2 ≠ 0, поэтому 3sinx+1 = 0

sinx = -1/3

x = 2pi*n + arcsin(-1/3) и x = 2pi*n + pi - arcsin(-1/3) (n - целое)

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7tg^2x + 2tgx - 5 = 0

(7tgx-5)(tgx+1) = 0

1) tgx = -1, x = -1/4pi + pi*n (n - целое)

2) tgx = 5/7, x = arctan(5/7) + pi*n (n - целое)

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2cos^2x - cosx - 3 = 0

(2cosx -3)(cosx + 1) = 0

1) cosx = -1, x = pi + 2pi*n (n - целое)

2) cosx = 3/2, невозможно, т.к. cos(x) ≤ 1

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2sinx = 1

sinx = 1/2

x = 1/6pi + 2pi*n и x = 5/6pi + 2pi*n (n - целое)