Решите неравенства: 1) 2sin²x-3√2sinx+2>0; 2) 4sin²x+5sinx+2cos²x>0. Created by alexclark169. algebra-ru

Решите неравенства: 1) 2sin²x-3√2sinx+2>0; 2) 4sin²x+5sinx+2cos²x>0...

July 2022 1 11 Report

Решите неравенства: 1) 2sin²x-3√2sinx+2>0;
2) 4sin²x+5sinx+2cos²x>0

Answers & Comments

sedinalana 1) 2sin²x-3√2sinx+2>0
sinx=t
2t²-3√2t+2>0
D=18-16=2
t1=(3√2-√2)/4=√2/2
t2=(3√2+√2)/4=√2
t<√2/2⇒sinx<√2/2⇒x∈(3π/4+2πk;9π/4+2πk.k∈z)
t>√2⇒sinx>√2 нет решения

2) 4sin²x+5sinx+2cos²x>0
4sin²x+5sinx+2-2sin²x>0
2sin²x+5sinx+2>0
sinx=t
2t²+5t+2>0
D=25-16=9
t1=(-5-3)/4=-2
t2=(-5+3)/4=-1/2
t<-2⇒sinx<-2 нет решения
t>-1/2⇒sinx>-1/2⇒x∈(-π/6+2πk;7π/6+2πk,k∈z)

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alexclark169 А почему в 1) "t<√2/2⇒sinx<√2/2⇒x∈(3π/4+2πk;9π/4+2πk.k∈z)", там разве не t>√2/2, sinx>√2/2?

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