Sin2x=2cosx*sinx
sin²x+cos²x=1
Поскольку х∈(0; π/2), то sinx=4/5
sin2x=2*4/5*3/5=2*4*3/25=24/25
x^4-17x^2+16 = 0
t^2-17t+16 = 0
t= 16
t = 1
x^2 = 16
x^2 = 1
x = 4
x = –4
x = 1
x = –1
X^4+9x^2-22≥0
x²=t
t²+9t-22≥0
D=9²-4*1*(-22)=81+88=169=13²
t1=(-9+13)/2=4/2=2
t2=(-9-13)/2=-22/2=-11
(t-2)(t+11)≥0
(x²-2)(x²+11)≥0
x²≥2
x1≤-√2
x2≥√2
x∈(-∞;-√2]U[√2;+∞)
См скриншоты
Ответ
≈ 0.77092
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