Question

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Answer 1

1
1)3x=+-π/6+2πn⇒x=+-π/18+2πn/3
2)x+π/4=πn⇒x=-π/4+πn
3)sinx=a
2a²-3a+1=0
D=9-8=1
a1=(3-1)/4=1/2⇒sinx=1/2⇒x=(-1)^n*π/6+πn
a2=(3+1)/4=1⇒sinx=1⇒x=π/2+2πn
2
1)3cos²x-2,5sinxcosx-2sin²x=0/cos²x≠0
2tg²x+2,5tgx-3=0
tgx=a
2a²+2,5a-3=0
D=6,25+24=30,25
a1=(-2,5-5,5)/4=-2⇒tgx=-2⇒x=-arctg2+πn
a2=(-2,5+5,5)/4=0,75⇒tgx=0,75⇒x=arctg0,75+πn
2)2√3xin(x/2)cos(x/2)-cos²(x/2)+sin²(x/2)-2sin²(x/2)-2cos²(x/2)=0
sin²(x/2)-2√3sin(x/2)cos(x/2)+3cos²(x/2)=0/cos²(x/2)≠0
tg²(x/2)-2√3tg(x/2)+3=0
(tgx/2-√3)²=0
tgx/2=√3
x/2=π/3+πn
x=2π/3+2πn
3)1+cos2x=cos²2x+2cos2xsin2x+sin²2x
1+cos2x=1+2sin2xcos2x
cos2x-2sin2xcos2x=0
cos2x(1-2sin2x)=0
cos2x=π/2+πn⇒x=π/4+πn/2
sin2x=1/2⇒2x=(-1)^n*π/6+πn⇒x=(-1)^n*π/12+πn/2
4)1-cosx≠0⇒cosx≠1⇒x≠2πn
-2sinxcos2x=0
sinx=0⇒x=πn+ОДЗ⇒x=π+2πn
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2
дополнительно
√(x³+8)=6
x³+8≥0⇒x³≥-8⇒x≥-2
x³+8=36
x³=24
x=∛24
Если 2√(x³+8)=6⇒√(x³+8)=3⇒x³+8=9⇒x³=1⇒x=1

sin²x+1,5sin2x-1=0
sin²x+3sinxcosx-cos²x-sin²x=0/cos²x≠0
cosx(3sinx-cosx)=0
cosx=0⇒x=π/2+πn
3sinx-cosx=0/cosx≠0
3tgx=1
tgx=1/3
x=arctg1/3+πn