Задача 59539 sin^4 x <0.5+cos2x ...

Адилет_376410300

3 мая 2021 г. в 12:23

sin⁴ x <0.5+cos2x

exponenta

3 мая 2021 г. в 12:41

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sin²x=(1–cos2x)/2

sin⁴x=(1–cos2x)²/4

(1–cos2x)²/4<0,5+cos2x

1–2cos2x+cos²2x <2+4cos2x
cos²2x–6cos2x–1<0

cos2x=t

D=36–4·=32

√32=4√2

t₁=(–6–4√2)/2; t₂=(–6+4√2)/2
t₁=–3–2√2; t₂=–3+2√2
⇒

–3–2√2<cos2x <–3+2√2

–(π–arccos(3–2√2)+2πk < 2x < π–arccos(3–2√2)+2πk, k ∈ Z

–(π/2)+(1/2)·arccos(3–2√2)+πk < x < (π/2)–(1/2) ·arccos(3–2√2)+πk, k ∈ Z