Как я понял, нужно решить 6 и 7.

В обоих номерах замена arcsin (13x/2) = t

6. t^2 - t = pi^2 - pi^2/4 = 3pi^2/4

D = 1 + 4*3pi^2/4 = 3pi^2 + 1

t1 = arcsin (13x/2) = (1 - √(3pi^2+1))/2

13x/2 = sin ((1 - √(3pi^2+1))/2)

[b]x1 = 2/13*sin ((1 - √(3pi^2+1))/2)[/b]

t2 = arcsin (13x/2) = (1 + √(3pi^2+1))/2

13x/2 = sin ((1 + √(3pi^2+1))/2)

[b]x2 = 2/13*sin ((1 + √(3pi^2+1))/2)[/b]

7. t^2 - pi/3*t = 0

t(t - pi/3) = 0

t1 = arcsin (13x/2) = 0

13x/2 = sin 0 = 0

[b]x1 = 0[/b]

t2 = arcsin (13x/2) = pi/3

13x/2 = sin (pi/3) = √3/2

[b]x2 = √3/13[/b]