Question

1. 3sin^2(2x)+7cos(2x)=3 2. cos^4(x/2)-cos^2(x/2)=0 3. ctg(3x/2)=1/√3 4. 1-ctg(4x-4pi)=0 5. ctg(4pi/√x)=-√3

Answer 1

1.$3sin ^{2} 2x+7cos2x=3$
ОДЗ: x∈R
$3sin ^{2}2x+7cos2x-3=0$
$3sin ^{2} 2x+7cos2x-3cos ^{2} 2x-3sin ^{2} 2x=0$
$-3cos ^{2} 2x+7cos2x=0$
$cos2x(3cos2x+7)=0$
[$cos2x=0$ ;            [x=$\pi /4+ \pi n/2; n$∈Z
[$3cos2x+7=0$;       [x=$\pi -arccos7/3+ \pi n;$ n∈Z

2. cos^4(x/2)-cos^2(x/2)=0
ОДЗ: х∈R
$cos ^{2} x/2(cos ^{2} x/2-1)=0$
cos²x/2=0;    1+сosx/2=0;  x= П+2Пn; n∈Z
cos²x/2=1;  1+сosx/2=1;    x=2Пn; n∈Z

3. ctg(3x/2)=1/√3
ОДЗ:х≠2Пn/3;n∈Z
3х/2=П/3+Пn; n∈Z
3x=2П/3+2Пn; n∈Z
х=2П/9+2Пn/3; n∈Z

4.1-ctg(4x-4pi)=0
ОДЗ: x≠5Пn/4;n∈Z
1+ctg4x=0
4х= -П/4+Пn;n∈Z
х= -П/16+Пn/4;n∈Z