Question

Решите уравнение:
9 arccos²x - 9π arccosx + 2π² = 0

Answer 1

-1 <= x <= 1

t = arccos x, 0 <= x <= П

9t^2 - 9Пt + 2П^2 = 0

t = 1/18*[9П +- √(81П^2 - 72П^2)] = П/18 * (9 +- 3) = {П/3, 2П/3}

x = cos arccos x = {cosП/3, cos2П/3} = +- 1/2