6sin²x - sinx - 1 = 0
Пусть sinx=t ( -1 <= t <= 1), тогда<br>
6t² - t - 1 = 0
D = 1 + 24 = 25 = 5^2
t₁ = (1 + 5)/12 = 1/2
t₂ = (1 - 5)/12 = - 1/3
Обратно возвращаемся
sinx = 1/2
x = (-1)^k*π/6 + πn, n ∈ Z
sinx = - 1/3
x = (-1)^(k+1)*arcsin(1/3) + πn, n ∈ Z