1) 4sinx > 2
sinx > 1/2
arcsin(1/2) + 2πn < x < π - arcsin(1/2) + 2πn, n∈Z
π/6 + 2πn < x < π - π/6 + 2πn, n∈Z
π/6 + 2πn < x < 5π/6 + 2πn, n∈Z
2) 2tg 2x+ 8> 0
tg2x > - 4
- arctg4 + πk < 2x < π/2 + πk, k∈Z
-(1/2)*arctg4 + (πk)/2 < x < π/4 + (πk)/2, k∈Z,
3) 5cos 2 x+2<7
5 cos2x < 5
cosx < 1
arccos(1) + 2πm < 2x < 2π - arccos(1) + 2πm, m∈Z
0 + 2πm < 2x < 2π - 0 + 2πm, m∈Z
2πm < 2x < 2π + 2πm, m∈Z
πm < x <π + πm, m∈Z
Task/24817273
---.---.---.--- .---.---
Найдите нули функции y = y =x⁴ - <span>2x² - 3
------------------
Решение : </span>
y =x⁴ - 2x² - 3 = x⁴ -3x² +x² - 3 =x²(x² - 3) +(x² - 3) =(x² - 3)(x² +1) =
(x+√3)(x -√3)(<span>x² +1) .
</span><span>y =0 ;
</span>(x+√3)(x -√3)(x² +1) =0 * * * x² +1 ≥ 1 ≠<span> 0 * * *
</span>x₁ = - √3 ;
x₂ = √3 .
ответ : ± <span>√3 .</span>
0,9:1+1/8=9/10+1/8=36/40+5/40=41/40=1 1/40=1,025
100% - х
80% - 940 р
х=(940*100)/80=1175 р