2cos²x(-(π/2+x)+√3sin2x=0
2sin²x+2√3sinxcosx=0
2sinx(sinx+√3cosx)=0
1) 2sinx=0
sinx=0
x=πn
2) sinx+√3cosx=0 |·1/2
1/2·sinx+√3/2·cosx=0
cosπ/3·six+sinπ/3·cosx=0
sin(π/3+x)=0
π/3+x=πn, n∈Z
x=πn-π/3, n∈Z
4*(0,4+( 6,4:4)-2
1)6.4 :4 = 1.6
2)0.4 +1.6 =2
3) 4 *2 = 8
<span>4) 8-2 =6</span>
Нет не имеет
4 5 1. 8 2. 6 7 3
7*7:(16-9)+(13+68):9*(36:4)