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tonya8
@tonya8
August 2022
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1) 2 cos (x-П/4)=(корень из 2 -2sinx)sin x
2) 2sin^2 x+2 sin(x+П/3)=sinx-1
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sedinalana
Verified answer
1
2сosx*cosπ/4+2sinx*sinπ/4-2sinx+2sin²x=0
2*√2/2*cosx+2*√2/2(sinx-√2sinx+2(1-cos²x)=0
√2cosx+√2sinx-√2sinx+2-2cos²x=0
2cos²x-√2cosx-2=0
cosx=a
2a²-√2a-2=0
D=2+16=18
a1=(√2-3√2)/4=-√2/2⇒cosx=-√2/2⇒x=+-3π/4+2πk,k∈z
a2=(√2+3√2)/4=√2⇒cosx=√2>1 нет решения
2
2(1-сos²x)+2sinxcosπ/3+2sinπ/3cosx-sinx+1=0
2-2cos²x+2*1/2*sinx+2*√3/2*cosx-sinx+1=0
3-2cos²x+sinx+√3cosx-sinx=0
2cos²x-√3cosx-3=0
cosx=a
2a²-√3a-3=0
D=3+24=27
a1=(√3-3√3)/4=-√3/2⇒cosx=-√3/2⇒x=+-5π/6+2πk,k∈z
a2=(√3+3√3)/4=√3⇒cosx=√3>1 нет решения
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Answers & Comments
Verified answer
12сosx*cosπ/4+2sinx*sinπ/4-2sinx+2sin²x=0
2*√2/2*cosx+2*√2/2(sinx-√2sinx+2(1-cos²x)=0
√2cosx+√2sinx-√2sinx+2-2cos²x=0
2cos²x-√2cosx-2=0
cosx=a
2a²-√2a-2=0
D=2+16=18
a1=(√2-3√2)/4=-√2/2⇒cosx=-√2/2⇒x=+-3π/4+2πk,k∈z
a2=(√2+3√2)/4=√2⇒cosx=√2>1 нет решения
2
2(1-сos²x)+2sinxcosπ/3+2sinπ/3cosx-sinx+1=0
2-2cos²x+2*1/2*sinx+2*√3/2*cosx-sinx+1=0
3-2cos²x+sinx+√3cosx-sinx=0
2cos²x-√3cosx-3=0
cosx=a
2a²-√3a-3=0
D=3+24=27
a1=(√3-3√3)/4=-√3/2⇒cosx=-√3/2⇒x=+-5π/6+2πk,k∈z
a2=(√3+3√3)/4=√3⇒cosx=√3>1 нет решения