1)7cos^2x-3 ctg^2x sin^2x если косинус=0,32)Решить уравнение sin^2*x/2-5sin*x/2+4=03)Найти наибольшее значение функции 5+2sinx.

а) f(х) = (sn (х) - cos (х) + х^2) = (sn (х)) (cos (х)) + (х^2) = cos (х) (-sn (х)) + 2 * х^1 = cos (х) + sn (х) + 2х.

б) f(х) = ((sn х) * (cos х 1)) = (sn х) * (cos х 1) + (sn х) * (cos х 1) = (sn х) * (cos х 1) + (sn х) * ((cos х) (1)) = (cos х)* (cos х 1) + (sn х) * ((-sn х) 0) = (cos^2 х) - (cos х) - (sn^2 х).

в) f(х) = (cos^2 (х / 3)) = (х / 3) * (cos (х / 3)) * (cos^2 (х / 3)) = (1 / 3) * (-sn (х / 3)) * 2 * (cos (х / 3)) = (2 / 3) * (-sn (х / 3)) * (cos (х / 3)).

г) f(х) = (sn^3 (2 - 3х)) = (2 - 3х) * (sn (2 - 3х)) * (sn^3 (2 - 3х)) = ((2) (3х)) * (sn (2 - 3х)) * (sn^3 (2 - 3х)) = (0 3) * (cos (2 - 3х)) * 3 * (sn^2 (2 - 3х)) = (-3) * (cos (2 - 3х)) * 3 * (sn^2 (2 - 3х)) = (-9) * (cos (2 - 3х)) * (sn^2 (2 - 3х).