Решение:
(2а-b)²=4a²-4ab+b²
Cos(π+x/2) =0 ;
π+x/2 =π/2+π*k , k∈ Z.
x/2 = - π/2 +π*k;
x= -π+2π*k;
x=(2k -1)π; нечетное число π
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sin(π/2+x) =1;
π/2+x =π/2 +2π*k;
x =2k*π. четное число π
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cos(π/3 -x) =√2/2;
cos(x-π/3) =√2/2;( cos четная фунлция)
x-π/3 =(+/-)π/4 +2π*
x₁= π/3+(+/-)π/4 +2π*k;
или
x₁ = 7π/12+2π*k;
x₂ = π/12+2π*k;
(x²-3x-10).√(4-x)=0
(x-5)(x+2).√(4-x)=0
a)x-5=0, x=5
b)x+2=0,x=-2
c)√(4-x)=0,4-x=0,x=4