Найти 0 функции у = log3(х2+5)−2 у = 4х - 3×2х – 4 у = √х2−4х+3 - id680554220210622 от Каракат11111 22.06.2021 06:51

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Алгебра: Найти 0 функции у = log3(х2+5)−2 у = 4х - 3×2х – 4 у = √х2−4х+3

Каракат11111 · Accepted Answer

1) 7tgx -10Ctgx +9 =0 |* tgx      7tg^2x -10 +9tgx = 0 tgx = y 7y^2 +9y -10 = 0 y1 = 10/14 = 5/7 у2 = -2 а) у = 5/7 tgx = 5/7 x = arctg5/7 + k, k ЄZ б) у = -2 tgx = -2 x = -arctg2 +k, k ЄZ 2) 10SinxCosx -14Cos^2x +2*1 = 0 10SinxCosx -14Cos^2x +2(Sin^2x+Cos^2x) = 0 10SinxCosx -14Cos^2x +2Sin^2x +2Cos^2x = 0 10SinxCosx -12Cos^2x +2Sin^2x = 0 :Cos^2x 10tgx -12 +2tg^2x= 0 tgx = y 2y^2 +10y -12=0 y^2 + 5y - 6 = 0 По т. Виета у1 = - 6 и  у2 = 1 а) у = - 6 tgx = -6 x = -arctg6+k, kЄZ  б)у = 1 tgx = 1 x...

Найти 0 функции у = log3(х2+5)−2 у = 4х - 3×2х – 4 у = √х2−4х+3

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