1) a) 2cosx-1=0
cosx=1/2
x=π/3+2πn, n∈Z
x=-π/3+2πn, n∈Z
б) cos²x+3sinx-3=0
1-sin²x+3sinx-3=0
-sin²x+3sinx-2=0
sinx=t
-t²+3t-2=0
D=9-8=1
t₁= (-3+1)/(-2)=1 t₂=(-3-1)/(-2)=2
sinx=1 sinx=2
x=π/2+2πn, n∈Z нет решения
в) 2sin²x-sin2x=cos2x
2sin²x-2sinxcos=cos²x-sin²x
3sin²x-2sinxcosx-cos²x=0 |÷sinx
3-2ctgx-ctg²x=0
ctgx=t
3t²-2t-1=0
D=4+12=16
t₁=(2+4)/-2=-3 t₂=(2-4)/-2=1
ctgx=-3 ctgx=1
x=arcctg(-3)+πn, n∈Z x=π/4+πn, n∈Z
x=π-arcctg3+πn, n∈Z
3) cos3x+cosx=0
2cos((3x+x)/2)cos((3x-x)/2 = 0
2cos2xcosx=0
cos2xcosx=0
cos2x=0 или cosx=0
2x=π/2+πn, n∈Z x=π/2+πn, n∈Z
x=π/4+πn/2, n∈Z
2. 8
3. 12
4. 15
5. 5
6. 11