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)
получим