(25^sinx)cos(2x)=5^sin(π-x))
5^(2sinx cos(2x)=5^sinx
2sinx cos(2x)=sinx
2sinx cos(2x)-sinx=0
sinx(2cos(2x)-1)=0
sinx=0 ili 2cos(2x)-1=0
x=πn cos(2x)=1/2
2x=+-π/3+2πn; n-celoe
x=+-π/6+πn
-5π/4 ≤ πn ≤ π/4; -5/4 ≤ n ≤ 1/4; n=-1;0
n=-1; x=-π;
n=0; x=0
2)-5π/4 ≤ -π/6+2πn ≤ π/4
-5π/4+π/6 ≤ 2πn ≤ π/4 +π/6
-13π/12 ≤2 πn ≤ 5π/12; -13/24 ≤ n ≤ 5/24; n= 0
n=0; x=-π/6
-5π/4 ≤ π/6+2πn ≤ π/4;
-5π/4-π/6 ≤ 2πn ≤ π/4-π/6; -17π/12 ≤ 2πn ≤ π/12; -8,5/12 ≤ n ≤1/24; n=0
n=0; x=π/6
Ответ. πn; +-π/6+2πn; n-celoe;-π;0;-π/6;π/6