3cos²x - 2,5sin2x - 2sin²x = 0 Разложим sin2x. 3cos²x - 5sinxcosx - 2sin²x = 0 Разделим на cos²x (cosx ≠ 0). 3 - 5tgx - 2tg² = 0 2tg²x + 5tgx - 3 = 0 Пусть t = tgx. 2t² + 5t - 3 = 0 D = 25 + 3•4•2 = 49 = 7². t = (-5 + 7)/4 = 1/2 t = (-5 - 7)/4 = -12/4 = -3 Обратная замена: tgx = 1/2 x = arctg(1/2) + πn, n ∈ Z tgx = -3 x = arctg(-3) + πn, n ∈ Z.
2) √3sinx - cosx = 2
√3/2sinx - 1/2cosx = 1 cos(π/6)sinx - sin(π/6)cosx = 1 По формуле синуса разности аргументов: sin(x - π/6) = 1 x - π/6 = π/2 + 2πn, n ∈ Z x = π/2 + π/6 + 2πn, n ∈ Z x = 2π/3 + 2πn, n ∈ Z.
Объяснение:
a) y'=( (4-3x)^2 )'= ( (4-3x)^2 )' * (4-3x)'=(-3)*2*(4-3x)=(-6)*(4-3x)=18x-24
б) y'=( (2-sqrt(x+2))^2 )'=( (2-sqrt(x+2))^2 )' * (sqrt(x+2))' * (x+2)'=2(2-sqrt(x+2))*(-1/(2sqrt(x+2)))*1= -(2-sqrt(x+2))/sqrt(x+2)=1-2/sqrt(x+2)
в) y'=( cosx*sqrt(2-x^2) )'=cosx' *sqrt(2-x^2)+cosx*(sqrt(2-x^2))' *(2-x^2)' = -sinx*sqrt(2-x^2)+cosx*(-2x)*1/(2sqrt(2-x^2))= -sinx*sqrt(2-x^2)-xcosx/sqrt(2-x^2)=(sinx(x^2-2)-xcosx)/sqrt(2-x^2)