2cosx-(2cos^2x-1)=0
2cosx-2cos^2x+1=0
-2cos^2x+2cosx+1=0|(-1)
2cos^2x-2cosx-1=0
cosx=t
2t^2-2t-1=0
t1=-(sqrt(3)-1)/2 t2=sqrt(3)/2+1/2
cosx=-(sqrt(3)-1)/2
x1=+-arccos-(sqrt(3)-1)/2+2piK
cosx=sqrt(3)/2+1/2
x=+-arccossqrt(3)/2+1/2+2piK
<span>Где к принадлежит Z
ТАК?
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(2/9-2.6)*4.5=2/9*9/2-2.6*4.5=1-11.7=-10.7
5x^3 - 20x = 0 //: 5
x^3 - 4x = 0
x* (x^2 - 4) = 0
x* (x - 2)*( x +2) = 0
x = 0
x = 2
x = - 2
{2x+7>4x-8⇒4x-2x<7+8⇒2x<15⇒x<7,5
{10+4x>0⇒4x>-10⇒x>-2,5
x∈(-2,5;7,5)