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MaryaKH
@MaryaKH
July 2022
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Решите уравнение
cos2x=2√2 sin(3pi/2 +x)+2
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sedinalana
Verified answer
2cos²x-1+2√2cosx-2=0
cosx=a
2a²+2√2a-3=0
D=8+24=32
a1=(-2√2-4√2)/4=-1,5√2⇒cosx=-1,5√2<-1 нет решения
f2=(-2√2+4√2)/4=√2/2⇒cosx-√2/2⇒x=+-π/4+2πn,n∈z
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Answers & Comments
Verified answer
2cos²x-1+2√2cosx-2=0cosx=a
2a²+2√2a-3=0
D=8+24=32
a1=(-2√2-4√2)/4=-1,5√2⇒cosx=-1,5√2<-1 нет решения
f2=(-2√2+4√2)/4=√2/2⇒cosx-√2/2⇒x=+-π/4+2πn,n∈z