Sin2x=cos(π/2 -x)
2sinx cosx=-sin(-x)
2sinx cosx=sinx
2sinx cosx - sinx=0
sinx (2cosx-1)=0
sinx=0 2cosx-1=0
x=πn, n∈Z 2cosx=1
cosx=1/2
x=+ π/3 +2πn, n∈Z
На промежутке [-π; 0]:
1) х=πn -π ≤ πn ≤ 0
-1 ≤ n ≤ 0
n=-1
x=-π
2) x= π/3 + 2πn -π≤ π/3 +2πn ≤ 0
-π-π/3 ≤ 2πn ≤ -π/3
-4π/3 ≤2πn ≤ -π/3
-2/3 ≤ n ≤ - 1/6
-4/6 ≤ n ≤ -1/6
здесь нет корней.
3) х=-π/3 + 2πn -π≤-π/3+2πn≤0
-π+π/3≤2πn≤π/3
-2π/3≤2πn≤π/3
-1/3≤ n ≤ 1/6
n=0
x= -π/3
Ответ: х=-π/3;
х=-π.