1) 10 sin 2x + 29 sin x-29 cos x = 31

2) 6sin^2 x-3sinx cosx-cos^2 x=1

3) 3tg^2 x-4cos^2 x=8

4) 5 sin 2x-12 (sin x-cos x) + 12=0

Таки решил!

10sin 2x + 29sin x - 29cos x = 31

20sin x*cos x + 29sin x - 29cos x = 31

29 (sin x - cos x) = 31 - 20sin x*cos x

29 (sin x - cos x) = 10 (sin^2 x + cos^2 x) - 20sin x*cos x + 21

29 (sin x - cos x) = 10 (sin x - cos x) ^2 + 21

Замена sin x - cos x = y

10y^2 - 29y + 21 = 0

D = 29^2 - 4*10*21 = 841 - 840 = 1

y1 = sin x - cos x = (29 - 1) / 20 = 28/20 = 7/5

y2 = sin x - cos x = (29 + 1) / 20 = 30/20 = 3/2

Есть формула

sin x - cos x = √2 * (1/√2*sin x - 1/√2*cos x) =

= √2 * (sin x*cos (pi/4) - cos x*sin (pi/4)) = √2*sin (x - pi/4)

Находим х

a) sin x - cos x = √2*sin (x - pi/4) = 7/5

sin (x - pi/4) = 7 / (5√2) = 7√2/10 ~ 0,98995 < 1

x - pi/4 = (-1) ^n*arcsin (7√2/10) + pi*n

x = pi/4 + (-1) ^n*arcsin (7√2/10) + pi*n

b) sin x - cos x = √2*sin (x - pi/4) = 3/2

sin (x - pi/4) = 3 / (2√2) = 3√2/4 ~ 1,06066 > 1

Решений нет

Ответ: x = pi/4 + (-1) ^n*arcsin (7√2/10) + pi*n

2) 6sin^2 x - 3sin x*cos x - cos^2 x = 1 = sin^2 x + cos^2 x

5sin^2 x - 3sin x*cos x - 2cos^2 x = 0

Делим все на cos^2 x

5tg^2 x - 3tg x - 2 = 0

Квадратное уравнение относительно tg x

(tg x - 1) (5tg x + 2) = 0

a) tg x = 1; x = pi/4 + pi*k

b) tg x = - 2/5; x = - arctg (2/5) + pi*k

3) 3tg^2 x - 4cos^2 x = 8

3sin^2 x / cos^2 x - 4cos^2 x - 8 = 0

Умножаем все на cos^2 x

3sin^2 x - 4cos^4 x - 8cos^2 x = 0

3 - 3cos^2 x - 4cos^4 x - 8cos^2 x = 0

Замена cos^2 x = y, 0 < = y < = 1 при любом х

4y^2 + 11y - 3 = 0

(y + 3) (4y - 1) = 0

y1 = cos^2 x = - 3 < 0 - решений нет

y2 = cos^2 x = 1/4

a) cos x = - 1/2; x1 = 2pi/3 + 2pi*k; x2 = 4pi/3 + 2pi*k

b) cos x = 1/2; x3 = pi/3 + 2pi*n; x4 = - pi/3 + 2pi*n

4) 5sin 2x - 12 (sin x - cos x) + 12 = 0

Решается также, как 1)

10sin x*cos x + 12 = 12 (sin x - cos x)

10sin x*cos x - 5 + 17 = 12 (sin x - cos x)

- (5sin^2 x + 5cos^2 x - 10sin x*cos x) + 17 = 12 (sin x - cos x)

-5 (sin x - cos x) ^2 + 17 = 12 (sin x - cos x)

Замена sin x - cos x = y.

-5y^2 + 17 = 12y

5y^2 + 12y - 17 = 0

(y - 1) (5y + 17) = 0

y1 = sin x - cos x = √2*sin (x - pi/4) = 1

sin (x - pi/4) = 1/√2

x1 - pi/4 = pi/4 + 2pi*n; x1 = pi/2 + 2pi*n

x2 - pi/4 = 3pi/4 + 2pi*n; x2 = pi + 2pi*n

y2 = sin x - cos x = √2*sin (x - pi/4) = - 17/5

sin (x - pi/4) = - 17 / (5√2) = - 17√2/10 ~ - 2,404 < - 1

Решений нет

Ответ: x1 = pi/2 + 2pi*n; x2 = pi + 2pi*n